= Solution
{c}
Direct <matrix multiplication> gives
$$
\begin{array}{c|ccc}
&X&Y&Z\\\hline
HPH^\dagger&Z&-Y&X\\
SPS^\dagger&Y&-X&Z.
\end{array}
$$
Thus 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 $P|\psi\rangle=\lambda|\psi\rangle$, then $(CPC^\dagger)C|\psi\rangle=\lambda C|\psi\rangle$. Therefore $C|\psi\rangle$ is another <Pauli eigenstate>. Every single-qubit Clifford circuit consequently permutes the six Pauli vertices of the <stabilizer octahedron>.
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