1. Let $A=\begin{pmatrix}a&b\\ b&c\end{pmatrix}$ since $A$ is a symmetric matrix of order $2$ with integer entries.

2. We need the sum of the diagonal elements of $A^2$.

   First compute $A^2$:

   =\begin{pmatrix}a^2+b^2 & ab+bc\\ ab+bc & b^2+c^2\end{pmatrix}.$$ Therefore, the sum of the diagonal elements of $A^2$ is $$a^2+b^2+b^2+c^2=a^2+2b^2+c^2.$$

3. Given that this sum is $1$, we have $a^2+2b^2+c^2=1.$

4. Since $a,b,c$ are integers, each term $a^2$, $2b^2$, $c^2$ is a non-negative integer. So we must solve $a^2+2b^2+c^2=1$ in integers.

5. Now check possibilities:

   • If $b\neq 0$, then $b^2\ge 1$, so $2b^2\ge 2$, making the left side at least $2$, impossible.
   • Hence $b=0$.

   Then the equation becomes $a^2+c^2=1.$

6. Integer solutions of $a^2+c^2=1$ are: $(a,c)=(\pm1,0),\ (0,\pm1).$

   Thus the corresponding matrices are

   \begin{pmatrix}-1&0\\0&0\end{pmatrix}, \begin{pmatrix}0&0\\0&1\end{pmatrix}, \begin{pmatrix}0&0\\0&-1\end{pmatrix}.$$

7. Therefore, the possible number of such matrices is $4.$

8. Option check:

   • A: $6$ ❌
   • B: $4$ ✅
   • C: $1$ ❌
   • D: $12$ ❌

Hence the correct answer is B.