I don't remember where I came across the statement but I'm pretty sure it is true and am interested in understanding why it holds. For any $n-$ register state $\rho^n \in H^{\otimes n}$ that is permutation invariant, there exists a purification that is also permutation invariant.
That is, if
$$\pi \rho^n = \rho^n$$
for any permutation $\pi$ among the $n$ registers, then there exists a purification $\vert\Psi^n\rangle\langle\Psi^n\vert \in (H\otimes H)^{\otimes n}$ of $\rho^n$ such that
$$(\pi\otimes \pi)\vert\Psi^n\rangle\langle\Psi^n\vert = \vert\Psi^n\rangle\langle\Psi^n\vert$$
for any permutation $\pi$. How does one see that this is true?
