I've read the following question:

Finite subgroups of ${\rm SL}_2(\mathbb{Z})$ (reference request)

and it made me wonder. It's easy to see that $\operatorname{SL}_2(\mathbb{Z})=\operatorname{Sp}_2(\mathbb{Z})$. So does it remain true that $\operatorname{Sp}_{2n}(\mathbb{Z})$ has only finitely many finite subgroups, if $n$ is general? If not, can we still say something about the possible orders of its finite subgroups? (For example, must they be divisible by only finitely many primes?)