Let us consider $n$ elements, each taken from the set $\{1, 2, \ldots, d\}$ and let $S_n$ be the set of all permutations on these $n$ elements.
Define a permutation operator on the set of $n$ qudits as
$$ P_d(\pi) = \sum_{i_1, i_2, \ldots, i_n \in [d]} |i_{\pi(1)}, \ldots, i_{\pi(n)}\rangle \langle i_1, \ldots i_n|. $$
Define the symmetric subspace on $n$ qudits as \begin{equation} V^{n}(\mathbb{C}^{d}) = \text{span}\{|\phi\rangle \in (\mathbb{C}^{d})^{\otimes n} : P_d(\pi)|\phi\rangle = |\phi\rangle~~\text{for all}~~\pi \in S_n \}. \end{equation}
Define an $n$ exchangeable density matrix $\rho \in \text{Density}\bigg(\big(\mathbb{C}^{d}\big)^{\otimes n}\bigg)$ as
$$ P_d(\pi) \rho P_d(\pi) = \rho, ~\text{for all} ~\pi \in S_n. $$
The notations are borrowed from here.
It is easy to see that any linear combination like
$$ \rho = \sum_i \alpha_i |\phi_i\rangle\langle \phi_i|, $$ where each $|\phi_i\rangle \in V^{n}(\mathbb{C}^{d})$ and the $\alpha_i$s are arbitrary complex numbers is an $n$ exchangeable state.
I could not prove the converse, however. Is it true that any $n$ exchangeable state is a linear combination of density matrices of states in the symmetric subspace? If not, what is an example to the contrary?