The short answer is: yes the inclusion in question is true for all maps $\Phi$ which are completely positive and completely co-positive.
In fact, it turns out to hold for the slightly more general scenario of completely positive (not necessarily trace-preserving) maps which are 2-co-positive (which obviously includes the completely co-positive case):
Theorem. Let $\Phi:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$ completely positive be given. If $\Phi$ is 2-co-positive, then ${\rm supp} (\Phi_c^* \circ \Phi_c)(|\psi\rangle\langle\psi|)\subseteq {\rm supp} (\Phi^* \circ \Phi)(|\psi\rangle\langle\psi|)$ for all $\psi\in\mathbb C^n$.
We will prove this in several steps. First of all let us see that the problem we're dealing with is unambiguous: While the complementary channel is highly non-unique as it depends on the chosen Kraus operators/Stinespring isometry, this non-uniqueness vanishes when considering the combination with its dual.
Lemma 1. For all completely positive maps $\Phi:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$ the map $\Phi_c^*\circ\Phi_c$ is independent of the original choice of $\Phi_c$. The Kraus operators of $\Phi_c^*\circ\Phi_c$ read $\{\Phi^*(|j\rangle\langle k|)\}_{j,k}$.
Proof. Start from any $V:\mathbb C^n\to\mathbb C^n\otimes\mathbb C^m$, $m\in\mathbb N$ such that $\Phi={\rm tr}_2(V(\cdot)V^\dagger)$. Thus the complementary channel $\Phi_c={\rm tr}_1(V(\cdot)V^\dagger)=\sum_j(\langle j|\otimes{\bf1})V(\cdot)V^\dagger(|j\rangle\otimes{\bf1})$ admits Kraus operators $\{(\langle j|\otimes{\bf1})V\}_j$ meaning its dual $\Phi_c^*$ has Kraus operators $\{((\langle j|\otimes{\bf1})V)^\dagger\}_j=\{V^\dagger(|j\rangle \otimes{\bf1})\}_j$. Combining these yields
\begin{align*}
\Phi_c^*\circ\Phi_c&=\sum_{k} \Phi_c^*\big((\langle k|\otimes{\bf1})V(\cdot)V^\dagger(|k\rangle\otimes{\bf1})\big)\\
&= \sum_{j,k} V^\dagger(|j\rangle \otimes{\bf1}) (\langle k|\otimes{\bf1})V(\cdot)V^\dagger(|k\rangle\otimes{\bf1})(\langle j|\otimes{\bf1})V\\
&= \sum_{j,k}V^\dagger(|j\rangle\langle k|\otimes{\bf1})V(\cdot)\big(V^\dagger(|j\rangle\langle k|\otimes{\bf1})V\big)^\dagger\,.
\end{align*}
The key insight now is that $V^\dagger((\cdot)\otimes{\bf1})V$ is precisely the dual of $\Phi$ because $$\Phi^*=( {\rm tr}_2\circ (V(\cdot)V^\dagger) )^*=(V(\cdot)V^\dagger)^*\circ{\rm tr}_2^*=V^\dagger{\rm tr}_2^*(\cdot)V=V^\dagger((\cdot)\otimes{\bf1})V\,.
$$
Therefore
$
\Phi_c^*\circ\Phi_c=\sum_{j,k}\Phi^*(|j\rangle\langle k|)(\cdot)(\Phi^*(|j\rangle\langle k|))^\dagger
$
is independent of the chosen $V$ and we even found the desired Kraus operators. $\square$
Now a quick sanity check to make sure that one in fact needs complete co-positivity and that the property in question is not already satisfied by all completely positive maps. In other words, however we will prove the statement in question we have to make use of complete (or some form of) co-positivity somehow.
Example. Consider the simplest completely positive map which is not (completely) co-positive: $\Phi={\rm id}=\Phi^*$. By Lemma 1 $\Phi_c^*\circ\Phi_c$ (is unique and) has Kraus operators $\{|j\rangle\langle k|\}_{j,k}$ which is known to imply that $\Phi_c^*\circ\Phi_c={\rm tr}(\cdot){\bf1}$. Thus for any $\psi\neq 0$
$$ {\rm supp} (\Phi_c^* \circ \Phi_c)(|\psi\rangle\langle\psi|)=\mathbb C^n\not\subseteq{\rm span}\{|\psi\rangle\}={\rm supp} (\Phi^* \circ \Phi)(|\psi\rangle\langle\psi|)$$
Often it is easier to work with the kernel (instead of the support) so let us take the orthogonal complement of ${\rm supp} (\Phi_c^* \circ \Phi_c)(|\psi\rangle\langle\psi|)\subseteq {\rm supp} (\Phi^* \circ \Phi)(|\psi\rangle\langle\psi|)$ to arrive at the equivalent $\!$ statement (because all involved operators are positive semi-definite, hence Hermitian) ${\rm ker} (\Phi_c^* \circ \Phi_c)(|\psi\rangle\langle\psi|)\supseteq {\rm ker} (\Phi^* \circ \Phi)(|\psi\rangle\langle\psi|)$.
The reason this helps us here is that we can now characterize when something is in the kernel.
Lemma 2. Given a linear map $\Phi:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$ and any $\psi,z\in\mathbb C^n$ the following statements hold.
- If $\Phi$ is positive, then $z\in {\rm ker} (\Phi^* \circ \Phi)(|\psi\rangle\langle\psi|)$ is equivalent to ${\rm tr}(\Phi(|z\rangle\langle z|)\Phi(|\psi\rangle\langle\psi|))=0$.
- If $\Phi$ is completely positive, then $z\in {\rm ker} (\Phi_c^* \circ \Phi_c)(|\psi\rangle\langle\psi|)$ is equivalent to $\Phi(|\psi\rangle\langle z|)=0$.
Proof. 1. Because $\Phi$ is positive, $(\Phi^* \circ \Phi)(|\psi\rangle\langle\psi|)\geq 0$ so $z$ is in its kernel if and only if $\langle z|(\Phi^* \circ \Phi)(|\psi\rangle\langle\psi|)|z\rangle=0$. But the latter can be expanded to
\begin{align*}
0&=\langle z|(\Phi^* \circ \Phi)(|\psi\rangle\langle\psi|)|z\rangle\\
&={\rm tr}(|z\rangle\langle z|(\Phi^* \circ \Phi)(|\psi\rangle\langle\psi|))\\
&={\rm tr}(\Phi(|z\rangle\langle z|)\Phi(|\psi\rangle\langle\psi|))
\,,
\end{align*}
as claimed.
2. As before $z\in {\rm ker} (\Phi_c^* \circ \Phi_c)(|\psi\rangle\langle\psi|)$ is equivalent to $\langle z|(\Phi_c^* \circ \Phi_c)(|\psi\rangle\langle\psi|)|z\rangle=0$. Using the Kraus operators of $\Phi_c^* \circ \Phi_c$ from Lemma 1 this yields
\begin{align*}
0=\langle z|(\Phi_c^* \circ \Phi_c)(|\psi\rangle\langle\psi|)|z\rangle&=\sum_{j,k}\langle z|\Phi^*(|j\rangle\langle k|)|\psi\rangle\langle\psi|\Phi^*(|j\rangle\langle k|)^\dagger|z\rangle\\
&=\sum_{j,k}|\langle z|\Phi^*(|j\rangle\langle k|)|\psi\rangle|^2
\end{align*}
so, equivalently, $\langle z|\Phi^*(|j\rangle\langle k|)|\psi\rangle=0$ for all $j,k$. But this we can re-write again:
$$
0=\langle z|\Phi^*(|j\rangle\langle k|)|\psi\rangle={\rm tr}(|\psi\rangle\langle z|\Phi^*(|j\rangle\langle k|))={\rm tr}(\Phi(|\psi\rangle\langle z|)|j\rangle\langle k|)=\langle k|\Phi(|\psi\rangle\langle z|)|j\rangle
$$
for all $j,k$ which is of course equivalent to $\Phi(|\psi\rangle\langle z|)=0$, as claimed. $\square$
The final lemma we need is one on positive semi-definite matrices.
Lemma 3. Given $A,B\in\mathbb C^{n\times n}$ positive semi-definite such that ${\rm tr}(AB)=0$ there exists $U\in\mathbb C^{n\times n}$ unitary and $c_1,\ldots,c_n\geq 0$ such that
\begin{align*}\begin{split}A&=U{\rm diag}(c_1,\ldots,c_r,0,\ldots,0)U^\dagger\\\text{and }\quad B&=U{\rm diag}(0,\ldots,0,c_{r+1},\ldots,c_n)U^\dagger\end{split}\tag{1}\end{align*}for some $r=0,\ldots,n$.
Moreover, if $C\in\mathbb C^{n\times n}$ is any matrix such that
$$
\begin{pmatrix}A&C\\C^\dagger&B\end{pmatrix}\geq 0\,,\tag{2}
$$
then there exists $C'\in\mathbb C^{r\times(n-r)}$ such that
$$
C=U\begin{pmatrix}0_r&C'\\0&0_{n-r}\end{pmatrix}U^\dagger\,.
$$
Proof. It is known that ${\rm tr}(AB)=0$ implies $AB=0$ and using cyclicity of the trace we also get $BA=0$. In particular this implies $AB=0=BA$ so because $A,B\geq 0$ they can be simultaneously unitarily diagonalized, i.e. there exist $W\in\mathbb C^{n\times n}$
unitary and $,a_{1},\ldots,a_{ n},b_{1},\ldots,b_{n}\geq 0$ such that
$A=W{\rm diag}(a_{1},\ldots,a_{ n})W^\dagger$ and $B=W{\rm diag}(b_{1},\ldots,b_{n})W^\dagger$.
Using this decomposition $AB=0$ is equivalent to $a_jb_j=0$ for all $j$, that is, for all $j=1,\ldots,n$ it holds that $a_j=0$ or $b_j=0$. This is almost what we want but we still have to group the zeros together: If $r=0,\ldots,n$ is the number of indices $j$ such that $b_j=0$ there certainly exists a permutation $\sigma$ such that $b_{\sigma(1)}=\ldots=b_{\sigma(r)}=0$ and $b_{\sigma(r+1)},\ldots$ are positive. It is now straightforward to verify that $U:=W\sigma^\dagger$ and $c_1:=a_{\sigma(1)},\ldots,c_r:=a_{\sigma(r)}$, $c_{r+1}:=b_{\sigma(r+1)},\ldots,c_n:=b_{\sigma(n)}$ satisfy (1).
For the additional statement assume (2) holds for some $C$. Then the claim follows at once the fact that positivity of the matrix
\begin{align*}
({\bf1}\otimes U)^\dagger\begin{pmatrix}A&C\\C^\dagger&B\end{pmatrix}({\bf1}\otimes U)&=\begin{pmatrix}U^\dagger A U&U^\dagger C U\\U^\dagger C^\dagger U&U^\dagger B U\end{pmatrix}\\&=
\begin{pmatrix}
\begin{matrix}{\rm diag}(c_1,\ldots,c_r)&0\\0&0
\end{matrix}&
\begin{matrix}C_{11}\ &\qquad C_{12}\qquad\quad\\C_{21}\ &\qquad C_{22}\qquad\quad
\end{matrix}\\
\begin{matrix}\qquad\quad C_{11}^\dagger\qquad&C_{21}^\dagger\\\qquad \quad C_{12}^\dagger\qquad&C_{22}^\dagger
\end{matrix}&
\begin{matrix}0&0\\0&{\rm diag}(c_{r+1},\ldots,c_n)
\end{matrix}
\end{pmatrix}
\end{align*}
forces that $C_{11},C_{21},C_{22}$ all vanish which yields the desired block structure with $C':=C_{12}$. $\square$
With this we have all the tools we need to prove the desired inclusion.
Proof of theorem. We will argue by contrapositive, that is, given any $\Phi$ completely positive, assuming there exists $\psi$ such that ${\rm supp} (\Phi_c^* \circ \Phi_c)(|\psi\rangle\langle\psi|)\not\subseteq {\rm supp} (\Phi^* \circ \Phi)(|\psi\rangle\langle\psi|)$ we will show that $\Phi\circ T$ is not 2-positive (i.e. $\Phi$ is not 2-co-positive). As seen above this assumption on the support is equivalent to $${\rm ker} (\Phi_c^* \circ \Phi_c)(|\psi\rangle\langle\psi|)\not\supseteq {\rm ker} (\Phi^* \circ \Phi)(|\psi\rangle\langle\psi|)$$ meaning there exists $z\in {\rm ker} (\Phi^* \circ \Phi)(|\psi\rangle\langle\psi|)$ such that $z\not\in {\rm ker} (\Phi_c^* \circ \Phi_c)(|\psi\rangle\langle\psi|)$.
By Lemma 2 this means that ${\rm tr}(\Phi(|\psi\rangle\langle\psi|)\Phi(|z\rangle\langle z|))=0$ but $\Phi(|\psi\rangle\langle z|)\neq 0$.
Next we have to draw a connection between these matrices: because $\Phi$ is completely positive $({\rm id}_2\otimes\Phi)(|\eta\rangle\langle\eta|)\geq 0$ for all $\eta\in\mathbb C^2\otimes\mathbb C^n$ so choosing $\eta:=|0\rangle\otimes|\psi\rangle+|1\rangle\otimes|z\rangle$ shows
$$
0\leq \begin{pmatrix}
\Phi(|\psi\rangle\langle\psi|)&\Phi(|\psi\rangle\langle z|)\\\Phi(|\psi\rangle\langle z|)^\dagger&\Phi(|z\rangle\langle z|)
\end{pmatrix}\,.
$$
This is where we apply Lemma 3 to get $U\in\mathbb C^{n\times n}$ unitary and $c_1,\ldots,c_n\geq 0$ such that
\begin{align*}\begin{split}
\Phi(|\psi\rangle\langle\psi|)&=U{\rm diag}(c_1,\ldots,c_r,0,\ldots,0)U^\dagger\,,\\
\Phi(|z\rangle\langle z|)&=U{\rm diag}(0,\ldots,0,c_{r+1},\ldots,c_n)U^\dagger\,,
\end{split}\tag{3}
\end{align*}
and $$\Phi(|\psi\rangle\langle z|)=U\begin{pmatrix}0_r&C'\\0&0_{n-r}\end{pmatrix}U^\dagger$$ for some $C'\in\mathbb C^{r\times(n-r)}$.
The key now is that $\Phi(|\psi\rangle\langle z|)\neq 0$ (because $z\not\in {\rm ker} (\Phi_c^* \circ \Phi_c)(|\psi\rangle\langle\psi|)$) meaning there exist $j=1,\ldots,r$, $k=r+1,\ldots,n$ such that
$$
\lambda:=(C')_{j(k-r)}=\langle j|U^\dagger \Phi(|\psi\rangle\langle z|)U|k\rangle\neq0\tag{4}
$$
(in particular this shows $r,n-r\geq 1$ because else $C'$ would be $0$ in the first place, a contradiction).
With this we are ready to construct $\xi,\xi'\in\mathbb C^2\otimes\mathbb C^n$ such that $\mu:=\langle\xi|({\rm id}_2\otimes(\Phi\circ T))(|\xi'\rangle\langle\xi'|)|\xi\rangle<0$ which would show that $\Phi\circ T$ is not 2-positive, as desired.
We choose
\begin{align*}
\xi&:=|0\rangle\otimes\sqrt\lambda^* U|k\rangle-|1\rangle\otimes\sqrt\lambda U|j\rangle\\
\text{and }\quad\xi'&:=|0\rangle\otimes|\overline\psi\rangle+|1\rangle\otimes|\overline z\rangle
\end{align*}
where $\sqrt\lambda\in\mathbb C$ is any square root of $\lambda$ (and $\overline{(\cdot)}$ denotes the entrywise complex conjugate $(\cdot)^*$). All that is left is to compute $\mu$: plugging in the definition of $\xi,\xi'$ and using the readily verified identity $|x\rangle\langle y|^T=|\overline y\rangle\langle\overline x|$ we find
\begin{align*}
\mu&=\Big\langle\begin{pmatrix} \sqrt\lambda^* U|k\rangle\\ -\sqrt\lambda U|j\rangle\end{pmatrix}\Big| \begin{pmatrix}\Phi(|\overline{\psi}\rangle\langle\overline{\psi}|^T)&\Phi(|\overline{\psi}\rangle\langle\overline{z}|^T)\\\Phi(|\overline{z}\rangle\langle\overline{\psi}|^T)&\Phi(|\overline{z}\rangle\langle\overline{z}|^T) \end{pmatrix} \Big| \begin{pmatrix} \sqrt\lambda^* U|k\rangle\\ -\sqrt\lambda U|j\rangle\end{pmatrix} \Big\rangle\\
&=\Big\langle\begin{pmatrix} \sqrt\lambda^* |k\rangle\\ -\sqrt\lambda |j\rangle\end{pmatrix}\Big| \begin{pmatrix}U^\dagger\Phi(|\psi\rangle\langle\psi|)U&U^\dagger\Phi(|z\rangle\langle\psi|)U\\U^\dagger\Phi(|\psi\rangle\langle z|)U&U^\dagger\Phi(|z\rangle\langle z|)U \end{pmatrix} \Big| \begin{pmatrix} \sqrt\lambda^* |k\rangle\\ -\sqrt\lambda |j\rangle\end{pmatrix} \Big\rangle\,.
\end{align*}
Two of the resulting four terms vanish as a result of Eq. (3): $\langle k|U^\dagger\Phi(|\psi\rangle\langle\psi|)U|k\rangle=0$ because $k>r$ and $\langle j|U^\dagger\Phi(|z\rangle\langle z|)U |j\rangle=0$ because $j<r+1$.
The two remaining terms then are easy to compute:
\begin{align*}
\mu&= \big\langle \sqrt\lambda^* \,k \,\big|U^\dagger\Phi(|z\rangle\langle\psi|)U\big| -\sqrt\lambda \,j\,\big\rangle+\big\langle -\sqrt\lambda \,j\,\big|U^\dagger\Phi(|\psi\rangle\langle z|)U\big| \sqrt\lambda^* \,k\, \big\rangle\\
&=-\lambda\langle k|U^\dagger\Phi(|z\rangle\langle\psi|)U|j\rangle-\lambda^* \langle j|U^\dagger\Phi(|\psi\rangle\langle z|)U|k\rangle\\
&=-\lambda(\langle j|U^\dagger\Phi(|\psi\rangle\langle z|)U|k\rangle)^*-\lambda^* \langle j|U^\dagger\Phi(|\psi\rangle\langle z|)U|k\rangle\\
&=-\lambda\lambda^*-\lambda^*\lambda=-2|\lambda|^2\overset{(4)}<0\tag*{$\square$}
\end{align*}
