Answer
No.
Work Step by Step
$\begin{bmatrix}
0& 1& 0 \\
1&0&0\\
0&0&1
\end{bmatrix} \begin{bmatrix}
0& 1&0 \\
1&0&0\\
0&0&1
\end{bmatrix}= \begin{bmatrix}
0\cdot0+1\cdot1+0\cdot0& ?& ? \\
?& ?& ?\\
?& ? & ?
\end{bmatrix}=\begin{bmatrix}
1& ?& ? \\
?&?&?\\
?&?&?
\end{bmatrix}$
We see that $A^2\ne A$.
Thus the answer is no.