For any positive integer $k$ coprime to $p$, and any positive integers $n,m$ there is a bijection between conjugacy classes of elements of order $k$ in ${\rm GL}_{n}(\mathbb{Z}/p^{m}\mathbb{Z})$ and conjugacy classes of elements of order $k$ in ${\rm GL}_{n}(\mathbb{Z}/p\mathbb{Z}).$ This is because ${\rm GL}_{n}(\mathbb{Z}/p^{m}\mathbb{Z})$ has a normal $p$-subgroup $U$ such that${\rm GL}_{n}(\mathbb{Z}/p^{m}\mathbb{Z})/U \cong {\rm GL}_{n}(\mathbb{Z}/p\mathbb{Z})$. ( It is also necessary to invoke the Schur-Zassenhaus Theorem).

Later edit: The matrix $A \in {\rm GL}_{n}(\mathbb{Z}/p\mathbb{Z})$ will have order $k$ if and only if the minimum polynomial $f(x)$ of $A$ is multiplicity free, and the lcm of the multiplicative order of the roots of $f(x)$ ( in the algebraic closure of $\mathbb{Z}/p\mathbb{Z}$) is $k$.

For any positive integer $k$ coprime to $p$, and any positive integers $n,m$ there is a bijection between conjugacy classes of elements of order $k$ in ${\rm GL}_{n}(\mathbb{Z}/p^{m}\mathbb{Z})$ and conjugacy classes of elements of order $k$ in ${\rm GL}_{n}(\mathbb{Z}/p\mathbb{Z}).$ This is because ${\rm GL}_{n}(\mathbb{Z}/p^{m}\mathbb{Z})$ has a normal $p$-subgroup $U$ such that${\rm GL}_{n}(\mathbb{Z}/p^{m}\mathbb{Z})/U \cong {\rm GL}_{n}(\mathbb{Z}/p\mathbb{Z})$. ( It is also necessary to invoke the Schur-Zassenhaus Theorem).

Later edit: The matrix $A \in {\rm GL}_{n}(\mathbb{Z}/p\mathbb{Z})$ will have order $k$ if and only if the minimum polynomial $f(x) \in \left(\mathbb{Z}/p\mathbb{Z}\right) [x]$ of $A$ is multiplicity free, and the lcm of the multiplicative order of the roots of $f(x)$ ( in the algebraic closure of $\mathbb{Z}/p\mathbb{Z}$) is $k$.

For any positive integer $k$ coprime to $p$, and any positive integers $n,m$ there is a bijection between conjugacy classes of elements of order $k$ in ${\rm GL}_{n}(\mathbb{Z}/p^{m}\mathbb{Z})$ and conjugacy classes of elements of order $k$ in ${\rm GL}_{n}(\mathbb{Z}/p\mathbb{Z}).$ This is because ${\rm GL}_{n}(\mathbb{Z}/p^{m}\mathbb{Z})$ has a normal $p$-subgroup $U$ such that${\rm GL}_{n}(\mathbb{Z}/p^{m}\mathbb{Z})/U \cong {\rm GL}_{n}(\mathbb{Z}/p\mathbb{Z})$. ( It is also necessary to invoke the Schur-Zassenhaus Theorem).

For any positive integer $k$ coprime to $p$, and any positive integers $n,m$ there is a bijection between conjugacy classes of elements of order $k$ in ${\rm GL}_{n}(\mathbb{Z}/p^{m}\mathbb{Z})$ and conjugacy classes of elements of order $k$ in ${\rm GL}_{n}(\mathbb{Z}/p\mathbb{Z}).$ This is because ${\rm GL}_{n}(\mathbb{Z}/p^{m}\mathbb{Z})$ has a normal $p$-subgroup $U$ such that${\rm GL}_{n}(\mathbb{Z}/p^{m}\mathbb{Z})/U \cong {\rm GL}_{n}(\mathbb{Z}/p\mathbb{Z})$. ( It is also necessary to invoke the Schur-Zassenhaus Theorem).

Later edit: The matrix $A \in {\rm GL}_{n}(\mathbb{Z}/p\mathbb{Z})$ will have order $k$ if and only if the minimum polynomial $f(x)$ of $A$ is multiplicity free, and the lcm of the multiplicative order of the roots of $f(x)$ ( in the algebraic closure of $\mathbb{Z}/p\mathbb{Z}$) is $k$.