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Let $ A,B $ be two Paulis with the same order, and neither of which is a multiple of the identity. Then there always exists some Clifford $ C $ such that $$ CAC^\dagger=B $$

Let $ A_1,A_2 $ be two Paulis which anticommute and square to the identity. Let $ B_1,B_2 $ be another pair of Paulis that anticommute and both square to the identity.

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

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Yes. An example of a circuit that does it is a generalized swap. To exchange the encoded qubit $A$ defined by logical observables $A_X = A_1, A_Z = A_2$ for encoded qubit $B$ defined by logical observables $B_X = B_1, B_Z = B_2$ you need to perform the encoded cnot sequence

$$ C = \text{SWAP}_{A,B} = \text{CNOT}_{A \rightarrow B} \cdot \text{CNOT}_{B \rightarrow A} \cdot \text{CNOT}_{A \rightarrow B} $$

Here is a way to decompose the generalized CNOT gates into one and two qubit gates. Let $P$ and $Q$ be commuting Pauli products, such as $P=A_X$ and $Q=B_Z$. Let $r$ be one of the qubits involved in $P$. From your question, it's clear you know how to find an operation $M_{P \rightarrow Z_r}$ that conjugates $P$ into a single qubit observable $Z_r$. Note that applying this mapping will change $Q$ into $Q^\prime = M_{P \rightarrow Z_r}^\dagger \cdot Q \cdot M_{P \rightarrow Z_r}$. However, we know $Q^\prime$ must commute with $Z_r$ because $Q$ commuted with $P$ and the mapping preserves commutation. From this we know that $Q^\prime$ cannot have a term $X_r$ or $Y_r$, though it may have $Z_r$. This allows us to apply the desired CNOT effect by simply iterating over the terms of $Q^\prime$ controlling them by $r$:

$$\text{CNOT}_{P \rightarrow Q} = M_{P \rightarrow Z_r}^\dagger \cdot \left( \prod_{t \in M_{P \rightarrow Z_r}^\dagger \cdot Q \cdot M_{P \rightarrow Z_r}} \text{Control}(Z_r, t) \right) \cdot M_{P \rightarrow Z_r}$$

where $\text{Control}(Z_r, Z_r)$ is just $Z_r$ and otherwise $\text{Control}(Z_r, t)$ is either a CX, CY, or CZ depending on if $t$ is a single-qubit X observable, single-qubit Y observable, or single-qubit Z observable.

Note that a full swap is slightly more specific than what you asked for. You wanted $A \rightarrow B$ but I also gave you $B \rightarrow A$. If you only want the former, you can achieve it with two generalized CNOTs instead of three.

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