# Permutation covariant channels and their Stinespring dilations

I am interested in a quantum channel from $$A^{\otimes n}$$ to $$B^{\otimes n}$$ denoted as $$N_{A^{\otimes n} \rightarrow B^{\otimes n}}(\cdot)$$. Let $$\pi(\cdot)$$ be a permutation operation among the $$n$$ registers. I am given that

$$N_{A^{\otimes n} \rightarrow B^{\otimes n}}\left(\pi_{A^{\otimes n}}(\cdot)\right) = \pi_{B^{\otimes n}} \left(N_{A^{\otimes n} \rightarrow B^{\otimes n}}(\cdot)\right)$$

The Stinespring dilation of $$N_{A^{\otimes n}\rightarrow B^{\otimes n}}(\cdot)$$ is an isometry $$V_{A^{\otimes n}\rightarrow B^{\otimes n}E}$$ such that $$\text{Tr}_E\left(V (\cdot) V^\dagger\right) = N(\cdot)$$

Does there exist a Stinespring dilation of $$N_{A^{\otimes n}\rightarrow B^{\otimes n}}$$ such that

$$V_{A^{\otimes n} \rightarrow B^{\otimes n}E}\pi_{A^{\otimes n}} = \pi_{B^{\otimes n}}V_{A^{\otimes n} \rightarrow B^{\otimes n}E}$$

It is clear that one can achieve this if permutation operations on the $$E$$ are also allowed but now we are only allowed to permute on $$B^n$$.

No, such a dilation might not exist. For a counter-example that can easily be generalized, let $$\Phi$$ be any non-isometric channel from $$A$$ to $$B$$, let $$n=2$$, and let $$N = \Phi\otimes \Phi$$.
By a non-isometric channel I mean one that does not look like $$\Phi(\cdot) = V \cdot V^{\dagger}$$ for some isometry $$V$$ from $$A$$ to $$B$$, and so it must be that $$\Phi$$ has Choi rank at least 2. Equivalently, there must be a Kraus representation $$\Phi(\rho) = \sum_{k=1}^r M_k \rho M_k^{\dagger}$$ where $$r\geq 2$$ and $$\{M_1,\ldots,M_r\}$$ are linearly independent. (We can take them to be orthogonal if we wish.)
We can easily obtain a Sinespring dilation of $$N = \Phi\otimes\Phi$$ by taking $$V = \sum_{1\leq j,k\leq r} M_j\otimes M_k \otimes v_{j,k}$$ for any orthonormal set $$\{v_{j,k}\}\subset E$$, and every Stinespring dilation of $$N$$ is expressible in this way.
The permutation-invariance property suggested by the question implies that we must have $$\sum_{1\leq j,k\leq r} M_j\otimes M_k \otimes v_{j,k}=\sum_{1\leq j,k\leq r} M_k\otimes M_j \otimes v_{j,k}$$ and therefore $$M_j\otimes M_k = M_k\otimes M_j$$ for all $$j,k$$ by the orthonormality of the set $$\{v_{j,k}\}$$. This, however, contradicts the linear independence of $$M_1$$ and $$M_2$$ (for instance).