Can the product of two nonsymmetric matrices be symmetric?

Try something simple first:

$$\begin{bmatrix}0&0\\1&0\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&1\end{bmatrix}\;.$$

More generally, if $A$ is any square real matrix, $AA^T$ is symmetric: the $(i,j)$-entry is the dot product of the $i$-th row of $A$ and the $j$-th column of $A^T$, and the $j$-th column of $A^T$ is the $j$-th row of $A$, so the $(i,j)$-th entry of $AA^T$ is the dot product of the $i$-th and $j$-th rows of $A$. The $(j,i)$-th entry of $AA^T$ is then the dot product of the $j$-th and $i$-th rows of $A$, which is of course the same.

This is not the only kind of example, however:

$$\begin{bmatrix}0&0&0\\1&0&0\\0&0&0\end{bmatrix}\begin{bmatrix}0&0&0\\0&0&0\\1&0&0\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$

As an alternative counterexample that isn't $AA^T$ or $A^TA$: Let $A$ be any non-symmetric, invertible matrix. Then the inverse matrix $A^{-1}$ will not be symmetric either (*). However, $A \cdot A^{-1} = A^{-1} \cdot A = I$, the identity matrix, which is of course symmetric.

(*) Proof: Suppose $A^{-1} = (A^{-1})^T$. Then $$A^{-1} = (A^{-1})^T = (A^T)^{-1},$$ hence $A = A^T$. But we assumed that $A$ was non-symmetric.

Here's a very simple example.

Take for example the following matrix $A = \begin{pmatrix}1 & 3 \\ 2 & 2 \end{pmatrix}$ and its transpose $A' = \begin{pmatrix}1 & 2 \\ 3 & 2 \end{pmatrix}$, the product of $$A*A' = \begin{pmatrix}10 & 8 \\ 8 & 8 \end{pmatrix}$$ which is symmetric. So this holds quite trivially.