This is only a hint, not an answer.

There is a simple characterization of semisimple matrices over finite fields. Namely, if $A\in M_n(F_q)$, its eigenvalues lie in $F_{q^m}$, $m=lcm(2,\dots,n)$, and there is $P\in GL_n(F_{q^m})$ such that $P^{-1}AP$ is a diagonal of Jordan blocks $\lambda_i I + N_i$, $i=1,\dots,s$. But it is easy to see that $(\lambda_i I+N_i)^{q^m}=\lambda_i^{q^m} I$ (note that $q^m \gt n$), so that $A$ is semisimple if and only if $A^{q^m}=A$.

Now you might want to start to study the possibilities for vector spaces $V\subset M_n(F_q)$ (or other subvarieties) such that $A^{q^m}=A$ for all $A\in V$.

This is only a hint, not an answer.

There is a simple characterization of semisimple matrices over finite fields. Namely, if $A\in M_n(F_q)$, its eigenvalues lie in $F_{q^m}$, $m=lcm(2,\dots,n)$, and there is $P\in GL_n(F_{q^m})$ such that $P^{-1}AP$ is a diagonal of Jordan blocks $\lambda_i I + N_i$, $i=1,\dots,s$. But it is easy to see that $(\lambda_i I+N_i)^{q^m}=\lambda_i I$ (note that $q^m \gt n$), so that $A$ is semisimple if and only if $A^{q^m}=A$.

Now you might want to start to study the possibilities for vector spaces $V\subset M_n(F_q)$ (or other subvarieties) such that $A^{q^m}=A$ for all $A\in V$.