The condition $Q^{-1}=Q^T$ means $Q$ is orthogonal, so its columns form an orthonormal set.

Write $Q=[q_{ij}]$. The commutation $PQ=QP$ forces equality of corresponding entries of $PQ$ and $QP$. Comparing, every off-diagonal entry connecting the eigenspace of $2$ (rows/columns $1,2$) and the eigenspace of $3$ (row/column $3$) must vanish. Hence

$q_{13}=q_{23}=q_{31}=q_{32}=0,$

so $Q$ has block form $Q=\left(\begin{array}{c}A0\\ 0c\end{array}\right)$, where $A$ is a $2\times 2$ orthogonal matrix with integer entries and $c\in \{-1,1\}$.

Integer orthogonal $2\times 2$ matrices have rows that are unit vectors with integer entries, namely $(\pm 1,0)$ or $(0,\pm 1)$. Counting such $A$: choose the first row in $4$ ways, then the second row must be perpendicular and unit, giving $2$ ways. So there are $8$ choices for $A$.

Combined with $2$ choices for $c$, the total is $8\times 2=16$.

$\therefore$ The correct option is (C).