Consider a quantum channel $\Phi : M_n \rightarrow M_m$ and let $\frac{\mathbb{I}_n}{n}$ be the maximally mixed input state. For all input states $\rho\in M_n$, is it true that
$$\quad \text{rank} \, \Phi (\rho) \leq \text{rank}\, \Phi \left(\mathbb{\frac{I_n}{n}}\right)?$$
Let's forget about normalization of the two sides of the inequality as if $\operatorname{rank}(A) \leq \mathrm{rank}(B)$ then $\mathrm{rank}(c A) \leq \mathrm{rank}(d B)$ for any $c,d \neq 0$.
Now let's write $I_n = \rho + (I-\rho)$. As $\rho \leq I_n$ because $\rho$ is a quantum state, we have $(I-\rho) \geq 0$. Then $$ \begin{aligned} \Phi(I_n) &= \Phi(\rho + (I_n-\rho)) \\ &= \Phi(\rho) + \Phi(I_n-\rho) \end{aligned} $$ Thus $\mathrm{rank}(\Phi(I_n)) = \mathrm{rank}(\Phi(\rho) + \Phi(I_n-\rho))$. But as $\Phi(\rho)$ and $\Phi(I_n - \rho)$ are both positive-semidefinite we have $\mathrm{rank}(\Phi(\rho) + \Phi(I_n-\rho)) \geq \mathrm{rank}(\Phi(\rho))$ (see this answer for a proof of that).
Putting it all together we get $$ \mathrm{rank}(\Phi(I_n)) \geq \mathrm{rank}(\Phi(\rho) ) $$ which implies the idenity in the question.
Regarding the title of the question You should not interpret this result as saying that the maximally mixed state is always mapped to a state of maximal rank in the output system. For example, you can take take a channel $\Phi(\rho) = \mathrm{Tr}(\rho) |0\rangle \langle 0 |$ whose output will always have rank $1$ regardless of the dimension of the output sysytem.
Yes, since for an arbitrary state $\rho \in M_n$, there exists a $\delta > 0$ such that $\delta \rho \leq \mathbb{I_n}/n$. The positivity of the channel then implies that $\Phi (\delta \rho) = \delta \Phi(\rho) \leq \Phi (\mathbb{I_n}/n)$, which clearly implies that $$ \text{rank}\, \Phi (\rho) \leq \text{rank}\, \Phi (\mathbb{I_n}/n).$$
Consider the Kraus representation of the channel $\Phi:\mathrm{Lin}(\mathcal X)\to\mathrm{Lin}(\mathcal Y)$: $$\Phi(X)=\sum_{a=1}^d A_a X A_a^\dagger,\tag1$$ for all $X\in\mathrm{Lin}(\mathcal X)$, for some positive integer $d$ and set of $d$ linear operators $A_a\in\mathrm{Lin}(\mathcal X,\mathcal Y)$ such that $\sum_a A_a^\dagger A_a=I_{\mathcal X}$.
Note that $\Phi(I_{\mathcal X}) = \sum_{a=1}^d A_a A_a^\dagger$. Therefore $$\operatorname{supp}(\Phi(I_{\mathcal X}))=\bigcup_a \operatorname{supp}(A_a),$$ and $\operatorname{supp}(\Phi(X))\subseteq \operatorname{supp}(\Phi(I_{\mathcal X}))$.
