Past exam of the introduction to quantum information course of the University of Oxford / 2021 / 1 / d

Direct matrix multiplication givesThus each generating Clifford gate conjugates every Pauli operator to another Pauli operator up to sign. Repeated conjugation proves the same for every product of generators and their inverses, hence for the Clifford group. For Hermitian Pauli observables the possible phases here are signs; arbitrary Pauli-group elements also include their conventional phases.

If $P|\psi ⟩=\lambda |\psi ⟩$, then $(CP{C}^{†})C|\psi ⟩=\lambda C|\psi ⟩$. Therefore $C|\psi ⟩$ is another Pauli eigenstate. Every single-qubit Clifford circuit consequently permutes the six Pauli vertices of the stabilizer octahedron.