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This a follow-up question to Conjugating pairs of Paulis to each other with a Clifford

We call a Clifford gate local if it is a tensor product of single qubit Clifford gates.

We call a Clifford gate non entangling if it is generated by single qubit Clifford gates and permutations.

Let $ A,B $ be two Paulis with the same weight $ wt(A)=wt(B) $. Then there always exists some non-entangling Clifford $ C $ such that $$ CAC^\dagger=B $$

Let $ A_1,A_2 $ be two Paulis which anticommute. Let $ B_1,B_2 $ be another pair of Paulis that anticommute. (Or you can assume that both pairs commute the answer shouldn't change.) Moreover suppose that $ wt(A_1)=wt(B_1) $ and $ wt(A_2)=wt(B_2) $.

Does there always exist some non-entangling Clifford $ C $ such that simultaneously $$ CA_1C^\dagger=B_1 $$ $$ CA_2C^\dagger=B_2 $$

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    $\begingroup$ Are you allowing permutations as well? $ZI$ and $IZ$ aren’t locally Clifford equivalent. $\endgroup$
    – squiggles
    Commented Apr 8, 2023 at 3:42
  • $\begingroup$ @squiggles great point! I should have said equivalent by local Cliffords + permutations (in other words equivalent by non-entangling Cliffords) $\endgroup$ Commented Apr 8, 2023 at 11:18

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