Proposition: Equivalence alone does not determine accessible recovery

Fix the same finite categorical equivalence and named label. Without a fixed positive encoding/access bridge, independent operational completions can have zero or nonzero full qubit recovery error.

Proof. For a fixed $k$, append equal-dimensional accessible and inaccessible coefficient spaces. One isometry sends

and another sends

Keep the local-system monodromy and categorical character decomposition unchanged. The first accessible channel is identity; the second is replacement. Their optimal qubit recovery errors are respectively zero and $3/4$, by the Bell argument in Theorem thm:loss.