Consider the matrix

$$P = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}.$$

Let the transpose of a matrix $$X$$ be denoted by $$X^T$$. Then the number of $$3 \times 3$$ invertible matrices $$Q$$ with integer entries, such that

$$Q^{-1} = Q^T \quad \text{and} \quad PQ = QP,$$

is

The two conditions on the integer matrix $$Q$$ are

1. $$Q^{-1}=Q^{T}$$    ⇒  $$Q$$ is an orthogonal matrix with integer entries.
2. $$PQ=QP$$ with $$P=\begin{pmatrix}2&0&0\\0&2&0\\0&0&3\end{pmatrix}$$.

Step 1: Characterise all $$3\times3$$ integer orthogonal matrices.
If $$Q$$ is orthogonal and its entries are integers, each row (and each column) is a unit vector whose components are integers. The only integer vectors of length 1 are the six vectors$$(\pm1,0,0),\;(0,\pm1,0),\;(0,0,\pm1).$$Hence every row and every column of $$Q$$ contains exactly one entry $$\pm1$$ and the rest are zeros. Such matrices are called signed permutation matrices.

The total number of signed permutation matrices of order 3 is $$3!\times2^{3}=6\times8=48$$ (choose a permutation of the columns and then choose the sign of the non-zero entry in each column).

Step 2: Impose the commutation condition $$PQ=QP$$.
Write $$Q=(q_{ij})$$. From $$PQ=QP$$ we get, for every pair $$(i,j),$$ $$P_{ii}\,q_{ij}=q_{ij}\,P_{jj}.$$
Because $$P=\text{diag}(2,2,3),$$ this gives $$(P_{ii}-P_{jj})\,q_{ij}=0\quad\forall\,i,j.$$ Whenever $$P_{ii}\neq P_{jj},$$ the factor $$P_{ii}-P_{jj}$$ is non-zero, forcing $$q_{ij}=0.$

Thus every entry mixing the first two coordinates with the third coordinate must vanish: $$q_{13}=q_{23}=q_{31}=q_{32}=0.$$ Therefore $$Q$$ has the block-diagonal form $$Q=$$\begin{pmatrix}$$ \ast & \ast & 0\\ \ast & \ast & 0\\ 0 & 0 & \ast \end{pmatrix},$$ where the asterisks are the remaining possible $$$$\pm$$1$$ or $$0$$ entries.

Step 3: Count the admissible signed permutation matrices that satisfy the block structure.
• The lower-right $$1$$\times$$1$$ block must be $$$$\pm$$1$$ (two choices).
• The upper-left $$2$$\times$$2$$ block must itself be a signed permutation matrix (to keep orthogonality and to make each column/row have exactly one non-zero entry).
  - There are $$2! = 2$$ ways to permute the two columns.
  - Independently, each of the two non-zero entries can be $$+1$$ or $$-1$$, giving $$2^{2}=4$$ sign choices.

Hence the number of possible $$2$$\times$$2$$ blocks is $$2$$\times$$4=8$$.
Multiplying by the two choices in the $$1$$\times$$1$$ block, we obtain the final count $$8$$\times$$2 = 16.$$

Step 4: Conclusion.
Exactly $$16$$ integer matrices $$Q$$ satisfy both $$Q^{-1}=Q^{T}$$ and $$PQ=QP$$.

Option C which is: $$16$$