Does the unitary freedom in the choice of Kraus operators for a given quantum channel just come from the unitary freedom in choice of purification of a quantum state?
Here's what I'm thinking. If I have two representations of the same quantum channel $$\Lambda(\rho) = \sum_k A_k\rho A_k^\dagger = \sum_k B_k \rho B_k^\dagger = \Lambda'(\rho),$$ then we also have $$(\Lambda \otimes id) \Omega = (\Lambda' \otimes id)\Omega,$$ where $\Omega = \sum_i \frac{1}{d}\vert{ii}\rangle\langle ii \vert.$ Each of these results in different representations of the same density matrix. So, they can be purified and there is unitary freedom in the choice of purification. Tracing out the auxiliary state seems to explain why there is unitary freedom in the choice of Kraus operators.
My question is twofold: is this correct? Is there a better (or another) way to see why we have unitary freedom in Kraus operators?