The Exercise

Let $S$ be a subgroup of the Pauli group $G_n$ on $n$ qubits that does not contain $-I$, and let $g$ be an element of $G_n$. The exercise asks to show that if conjugation by $g$ maps $S$ onto itself, possibly permuting its elements, then $g$ commutes with every element of $S$.

The Pauli Group

The Pauli group $G_n$ on $n$ qubits is the group, under multiplication, of the operators $$i^s\, P_1 \otimes \cdots \otimes P_n, \qquad s \in \mathbb{Z}_4, \quad P_j \in \{I, X, Y, Z\},$$(1)^✦ each given by its phase power $s$, an integer modulo 4, and its Pauli string, the choice of $P_j$ among the identity $I$ and the Pauli operators $X$, $Y$, $Z$ for each qubit $j$.

The element $-I$ of $G_n$ is the identity string with phase power $s = 2$, $$-I = i^2\, I \otimes \cdots \otimes I.$$(2)^✦

For a subgroup $S$ of $G_n$, the normalizer $N(S)$ and the centralizer $Z(S)$ of $S$ are the subgroups of $G_n$ $$N(S) = \{\, g \in G_n : g S g^\dagger = S \,\}, \qquad Z(S) = \{\, g \in G_n : g s = s g \text{ for all } s \in S \,\}.$$(3)

The Claim

What Is Checked

The claim is stated over the Pauli group $G_n$ of (1), on any number $n$ of qubits, and over its subgroups $S$.

About these constructions the library shows that $G_n$ is a group under multiplication, with $-I$ among its elements.