= Solution
Put $a=\lambda_P$ and $b=\lambda_Q$. The <Pauli operators> are Hermitian and satisfy $P^2=Q^2=I$ and $PQ+QP=0$, so
$$
V^\dagger=V,
\qquad
V^2=\frac12(aP+bQ)^2=I.
$$
Thus $V$ is a <unitary operator>. For every Pauli operator $R$, according as $R$ commutes or anticommutes with $P$ and $Q$, expansion of $VRV$ gives one of $\pm R$ or $\pm RPQ$, up to the phase that makes it Hermitian. It is therefore another Pauli operator, so $V$ normalizes the <Pauli group> and is a <Clifford operation>. Finally,
$$
V|\psi\rangle
=\frac{1}{\sqrt2}(I+bQ)|\psi\rangle,
$$
which is the normalized projection onto the eigenvalue-$b$ <eigenspace> of $Q$. Hence $V$ maps the eigenvalue-$\lambda_P$ eigenspace of $P$ onto the eigenvalue-$\lambda_Q$ eigenspace of $Q$.
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