TITLE: $GL_n(\Bbb Z_p)$ conjugacy classes in a $GL_n(\Bbb Q_p)$ conjugacy class
QUESTION [20 upvotes]: It is easy to classify conjugacy classes in $GL_n(\mathbb Q_p)$ by linear algebra. How to classify $GL_n(\Bbb Z_p)$ conjugacy classes in a $GL_n(\Bbb Q_p)$ conjugacy class? For example, for general matrix $A \in GL_n(\mathbb Z_p)$, how many $X \in GL_n(\mathbb Z_p)$ are  $GL_n(\mathbb Q_p)$ conjugated to $A$ up to $GL_n(\mathbb Z_p)$ conjugaction? In some cases it is finite, is it always finite?
Motivation: when counting invariant lattices, one comes across such problems.
Edit: could we find some numerical invariants that distinguish different $\Bbb Z_p-$ conjugacy classes?

REPLY [7 votes]: The answer is yes if $A \in \mathrm{GL}_n(\mathbb{Z}_p)$ is semisimple.
We may think of a matrix $A \in M_n(\mathbb{Z}_p)$ as a $\mathbb{Z}_p$-lattice of rank $n$ endowed with a $\mathbb{Z}_p$-linear endomorphism: take the standard lattice $L=\mathbb{Z}_p^n$ endowed with the endomorphism $\varphi$ associated to $A$. Clearly, two matrices $A,A' \in M_n(\mathbb{Z}_p)$ are $\mathrm{GL}_n(\mathbb{Z}_p)$-conjugated if and only if the associated pairs $(L,\varphi)$ and $(L',\varphi')$ are isomorphic.
More formally, consider the algebra $R = \mathbb{Z}_p[A]$ inside $M_n(\mathbb{Z}_p)$. Then $A$ gives rise to an $R$-module $M$ which is $\mathbb{Z}_p$-free of rank $n$. Note that if $A' \in M_n(\mathbb{Z}_p)$ is another matrix which is $\mathbb{Q}_p$-conjugated to $A$ then the algebras $\mathbb{Z}_p[A]$ and $\mathbb{Z}_p[A']$ are isomorphic, as they are both isomorphic to $R = \mathbb{Z}_p[X]/(\mu)$ where $\mu$ denotes the minimal polynomial. However, the associated $R$-modules $M$ and $M'$ need not be isomorphic. In fact, they are if and only if $A$ and $A'$ are $\mathbb{Z}_p$-conjugated.
So the problem reduces to classifying the $R$-modules which are $\mathbb{Z}_p$-free of rank $n$ up to isomorphism. In the semisimple case, the following theorem of Jordan-Zassenhaus gives a positive answer to your third question.

Theorem (Jordan-Zassenhaus). Let $K$ be a local or global field with ring of integers $\mathcal{O}_K$. Let $L$ be a (commutative) semisimple $K$-algebra, and let $R$ be an $\mathcal{O}_K$-order in $L$. Then for any integer $n \geq 1$, the set of isomorphism classes of $R$-modules which are $\mathcal{O}_K$-lattices of rank $\leq n$ is finite.

The case at hand follows by taking $K=\mathbb{Q}_p$, $L=\mathbb{Q}_p[X]/(\mu)$ and $R=\mathbb{Z}_p[X]/(\mu)$ where $\mu$ is a squarefree polynomial. A proof of the Jordan-Zassenhaus theorem is given in Reiner, Maximal orders, (26.4), p. 228. It is stated only for global fields, but you can check it also holds for local fields.
EDIT. The proof is not difficult: first one treats the case where $R$ is a maximal order, which is clear since it is a product of discrete valuation rings (hence PIDs). If $R' \subset R$ is an arbitrary order and $M'$ is an $R'$-lattice then there are only finitely many possibilities for $M=M' \otimes_{R'} R$, and since $p^N M \subset M' \subset M$ there are only finitely many possibilities for $M'$. So we get the finiteness.
Regarding the question of an explicit classification, you may be interested by Marseglia's article Computing the ideal class monoid of an order, where he gives algorithms to compute explicit representatives (in the global case).
EDIT 2. Another article giving an algorithm in the global case appeared on the arXiv today: https://arxiv.org/abs/1811.06190