Let $A$ be an element of $\mathcal{M}_n$ such that $A$ is periodic. Then $A^\mathrm{T}$ is also periodic. However, $AA^\mathrm{T}$ is periodic if and only if each nonzero entry of $A$ is $1$ and each row and column of $A$ has at most one nonzero entry.
Suppose $A=(a_{ij})$ and $a_{i_0j_0}\gt1$ for some $i_0$ and $j_0$. If $AA^\mathrm{T}=(b_{ij})$, then $b_{i_0i_0}\geq a_{i_0j_0}^2\gt1$. In particular this implies that the Euclidean norm of $AA^\mathrm{T}e_{i_0}$ is greater than $1$, where $e_{i_0}$ is the $i_0^\text{th}$ standard basis vector. Since $AA^\mathrm{T}$ is positive semidefinite, this implies that its spectral radius is greater than $1$, and therefore its powers have unbounded spectral radius.
Suppose that $a_{i_0j_0}\gt0$ and $a_{i_0j_1}\gt0$ for some $i_0$ and $j_0\neq j_1$. Then $b_{i_0i_0}\geq a_{i_0j_0}^2+a_{i_0j_1}^2\gt1$, which again implies that the powers of $AA^\mathrm{T}$ have unbounded spectral radius.
Suppose that $a_{i_0j_0}\gt0$ and $a_{i_1j_0}\gt0$ for some $i_0\neq i_1$ and $j_0$. Then $b_{i_0i_0}\geq a_{i_0j_0}^2\gt0$ and $b_{i_1i_0}\geq a_{i_1j_0}a_{i_0j_0}\gt0$. This implies that the Euclidean norm of $AA^\mathrm{T}e_{i_0}$ is at least $\sqrt{2}$. Thus, again, the powers of $AA^\mathrm{T}$ have unbounded spectral radius.
On the other hand, if each nonzero entry of $A$ is $1$ and each row and column of $A$ has at most one nonzero entry, then $AA^\mathrm{T}$ is idempotent.