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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?

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    $\begingroup$ Yes, this is true, and not hard to prove. There' s a proof in Renato Renner's PhD thesis. $\endgroup$ Jun 2, 2021 at 8:44
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    $\begingroup$ Extended @Mateus 's answer. Here is the thesis link: arxiv.org/pdf/quant-ph/0512258.pdf , see lemma 4.2.2 $\endgroup$ Jun 2, 2021 at 17:07

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