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? 



$GL_n(\Bbb Z_p)$ conjugacy classes in a $GL_n(\Bbb Q_p)$ conjugacy class
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*](https://arxiv.org/abs/1805.09671), 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>


