Non conjugate $p$-subgroups of $\mathrm{GL}_n(\mathbb Z)$ It is probably not that difficult but I can't find an example of two non-conjugate $p$-subgroups (of same order) of $\mathrm{GL}_n(\mathbb Z)$ ($n>1$).
 A: Let $A_n$ be the companion matrix to the $n$-th cyclotomic polinomial. Then if $p$ is a prime, the block diagonal matrices $$\begin{pmatrix}A_p\\&A_p\\&&\ddots\\&&&A_p\end{pmatrix},$$ with $p$ copies of $A_p$ along the diagonal, and $$\begin{pmatrix}A_p\\&I\\&&\ddots\\&&&I\end{pmatrix},$$ with one $A_p$ along the diagonal and the rest of the blocks $p\times p$ identity matrices, generate subgroups of order $p$ in $GL(p^2,\mathbb Z)$ which are not conjugate.
You can play games like this in various other ways.
 
The classification of modules over a cyclic group of prime order over the integers (which are finitely generated and torsion free as abelian groups) was worked out by [F. E. Diederichsen, Über die Ausreduktion ganzzahliger Gruppendarstellungen bei arithmetischer Äquivalenz, Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg vol. 14 (1938) pp. 357-412.]
and Reiner gave a simpler exposition in [Reiner, Irving. Integral representations of cyclic groups of prime order. Proc. Amer. Math. Soc. 8 (1957), 142--146. MR0083493 (18,717a)] (which you can get from the nice people at the AMS) The classification, plus a bit more of work (modulo the knowledge of the ideal class group of cyclotomic fields...), provides a complete answer to your question in the prime-order case. Reiner then classified integral representations of groups of order $p^2$; this is more involved, as there are infinitely many indecomposables.
