$\DeclareMathOperator\GL{GL}$Say $p$ is an odd prime, and take two matrices $A,B\in \GL\_n({\mathbb Z}\_p)$ of finite order $m$. Is it true that they are conjugate in $\GL\_n({\mathbb Z}\_p)$ if and only if their reductions mod $p$ are conjugate in $\GL\_n({\mathbb F}\_p)$?


**Edit:** Thank you all for the answers and a very insightful discussion! As the answer is **NO**, it is important for me to know whether conjugacy in $\GL\_n({\mathbb F}\_p)$ at least implies conjugacy in $\GL\_n({\mathbb Q}\_p)$. I cannot see this from the answers, so I think I will ask this as a separate question.



Conjugacy for $p$-adic matrices of finite order
I think I finally have a correct answer for arbitrary $p$.


As F. Ladisch [notes](https://mathoverflow.net/a/72183), $G=C\_{p^3}$ has only finitely many indecomposable modular representations. For the following argument, I will not only need infinitely many integral representations, but I will need them to occur at reasonably regular intervals, as the rank goes up. Luckily, they do. In "[Representations of Cyclic Groups in Rings of Integers, II](https://doi.org/10.2307/1970218)", Heller and Reiner exhibit indecomposable $\mathbb{Z}\_p[G]$-modules of rank $kp^3$ for arbitrary $k\in \mathbb{N}$.


Now, let us count the number of arbitrary $\mathbb{F}\_p[G]$-modules, respectively of arbitrary $\mathbb{Z}\_p[G]$-modules, of rank $np^3$. The former correspond to partitions of $np^3$ into summands that come from a fixed finite set of integers (possibly some repetitions), and their number is easily seen to be polynomial in $n$ (e.g. if the finite set is $A$, then a very crude upper bound on the number of partitions is $\prod\_{i\in A} np^3/i$).


On the other hand, even if we consider only direct sums of the above mentioned modules of rank $kp^3$, $k\in \mathbb{N}$, we get a number that is essentially the partition number of $n$, which is [exponential in $\sqrt{n}$](http://en.wikipedia.org/wiki/Partition_(number_theory)#Partition_function_formulas). So for sufficiently large $n$, we get much more $\mathbb{Z}\_p[G]$-modules of rank $np^3$ than $\mathbb{F}\_p[G]$-modules. This shows the even stronger claim that arbitrarily many non-conjugate matrices over $\mathbb{Z}\_p$ can reduce to conjugate $\mathbb{F}\_p$-matrices.


It is also worth remarking that for $G=C\_p$, it is true that two $\mathbb{Z}\_p[G]$-modules are isomorphic if and only if their reductions are. That's because the only indecomposable $\mathbb{Z}\_p[G]$-modules are the trivial module, the regular module and the $p-1$-dimensional augmentation ideal in the regular module (at the moment, I can't find a good reference, so I might edit it in later). For $G=C\_{p^2}$, one could also go through the finite classification ([Reiner - Integral Representations of Cyclic Groups of Order $p^2$](https://doi.org/10.2307/2041351)) and see whether this still holds.


