Let $P_n$ be the Pauli group on $n$ qubits defined by $P_n=\{i^k \sigma_1...\sigma_n: k=0, 1, 2 \; \mathrm{or}\;3, \sigma_j\in \{I, X, Y, Z\}\}$, where $I, X, Y, Z$ are Pauli matrices. The Clifford group $C_n$ is defined as the normalizer of $P_n$ in the unitary group as follows: $C_n=\{U\in U(2^n):UP_n U^\dagger=P_n\}$. Is the following true?
$\{U\in C_n:U\sigma=\pm\sigma U \;\forall\sigma\in P_n\}=P_n$.
Thank you.
Yes, if a Clifford operation $C$ conjugates every Pauli product $P$ into $\pm P$, then $C$ is a Pauli product.
First, show that $PAP^\dagger = \pm A$ for any Pauli products $A$ and $P$. In other words, paulis-implies-only-sign-change. This just comes down to all pairs of Pauli products commuting or anti-commuting.
Second, show only-sign-change-implies-paulis. Take a promised only-sign-change operation $C$ and compute $C B C^\dagger$ versus the generators $B \in \{X_1, Z_1, X_2, Z_2, \dots\}$. The sign changes you get for $X_1$ and $Z_1$ reveal which of the four pauli operators (IXYZ) is being to qubit 1 by $C$, and same for each other qubit index $k$. Together they give you a reconstructed Pauli product, which you know is equal to $C$ because this operation and $C$ do the same thing when conjugating any generator $B$.
