1. Interpret the conditions

We need the number of $3 \times 3$ invertible integer matrices $Q$ such that:

$Q^{-1} = Q^T$ and $PQ = QP,$ where $P = \begin{pmatrix}2&0&0\\0&2&0\\0&0&3\end{pmatrix}.$

The condition $Q^{-1}=Q^T$ means that $Q$ is an orthogonal matrix.

Since $Q$ also has integer entries, each column of $Q$ is an integer vector of length $1$.

2. Use orthogonality with integer entries

Let a column of $Q$ be $v = (a,b,c)^T$ with $a,b,c \in \mathbb{Z}$. Because $Q$ is orthogonal, each column has norm $1$:

$a^2+b^2+c^2=1.$

The only integer solutions are:

$v = \pm e_1,\ \pm e_2,\ \pm e_3,$

where $e_1,e_2,e_3$ are the standard basis vectors.

So $Q$ must be a signed permutation matrix.

The total number of such $3\times 3$ matrices is

$3! \cdot 2^3 = 48,$

but we must also impose $PQ=QP$.

3. Apply the commutation condition $PQ=QP$

Since $P=\operatorname{diag}(2,2,3),$ it has eigenspaces:

• eigenvalue $2$: span of $e_1,e_2$
• eigenvalue $3$: span of $e_3$

Any matrix commuting with $P$ must preserve these eigenspaces. So $Q$ cannot mix the third coordinate with the first two.

Hence $Q$ must have block form

$Q = \begin{pmatrix}A & 0\\0 & \pm 1\end{pmatrix},$

where $A$ is a $2\times 2$ integer orthogonal matrix.

4. Count $2\times 2$ integer orthogonal matrices

By the same argument, a $2\times 2$ integer orthogonal matrix must be a signed permutation matrix.

There are:

$2! \cdot 2^2 = 8$

such matrices.

They are exactly the matrices obtained by permuting $e_1,e_2$ and assigning signs.

5. Choose the $(3,3)$ entry

The third coordinate must map to itself up to sign, so the bottom-right entry can be either $+1$ or $-1$. That gives $2$ choices.

Therefore total number of such matrices is

$8 \cdot 2 = 16.$

6. Check options

So the correct option is:

$\boxed{16}$

which is Option C.