TITLE: Conjugacy for p-adic matrices of finite order II
QUESTION [12 upvotes]: Question: 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 if their reductions mod $p$ are conjugate in $GL_n({\mathbb F}_p)$ then they are conjugate in $GL_n({\mathbb Q}_p)$?

This is a follow-up to the question whether $GL_n({\mathbb F}_p)$-conjugacy implies $GL_n({\mathbb Z}_p)$-conjugacy, for which the answer is "No". 
For $p=2$ this is not true, although I would be very much interested to know whether conjugacy mod 4 implies conjugacy over ${\mathbb Q}_2$.
(Note that if the answer is "Yes", then by character theory any two representations $\rho, \rho': G\to GL_n({\mathbb Z}_p)$ that are equivalent mod $p$ are equivalent in $GL_n({\mathbb Q}_p)$.)

REPLY [3 votes]: Edit: Sorry, just realized this is in in the wrong direction. What was asked for was an example of matrices over $\mathbb Z_p$ which are conjugate over $\mathbb F_p$ but not over $\mathbb Q_p$; I swapped $\mathbb Q_p$ and $\mathbb F_p$, but then of course the assertion is clear anyway (I'll look at it again later).

After looking at Geoff's answer and §34C in "Curtis and Reiner: Methods of Representation Theory, Volume I" which deals with integral representations of $C_{p^2}$, I came up with a counterexample (I would recommend to use GAP or Maple to verify it; doing the computations by hand would be insane). 
So, here is the counterexample: Consider the following two matrices in $\mathbb Q_3^{10\times 10}$:
$$
A=\left(\begin{array}{rrrrrrrrrr}
1&0&0&0&1&0&0&0&0&0\newline
0&0&0&1&1&0&0&0&0&0\newline
0&1&0&0&1&0&0&0&0&0\newline
0&0&1&0&1&0&0&0&0&0\newline
0&0&0&0&0&0&0&0&0&-1\newline
0&0&0&0&1&0&0&0&0&0\newline
0&0&0&0&0&1&0&0&0&0\newline
0&0&0&0&0&0&1&0&0&-1\newline
0&0&0&0&0&0&0&1&0&0\newline
0&0&0&0&0&0&0&0&1&0\newline
\end{array}\right)
$$
and
$$
B = \left(\begin{array}{rrrrrrrrrr}%
1&0&0&0&0&0&0&0&0&0\newline
0&0&0&1&0&0&0&0&0&0\newline
0&1&0&0&0&0&0&0&0&0\newline
0&0&1&0&0&0&0&0&0&0\newline
0&0&0&0&0&0&0&0&0&-1\newline
0&0&0&0&1&0&0&0&0&0\newline
0&0&0&0&0&1&0&0&0&0\newline
0&0&0&0&0&0&1&0&0&-1\newline
0&0&0&0&0&0&0&1&0&0\newline
0&0&0&0&0&0&0&0&1&0\newline
\end{array}\right)
$$
Then $A$ and $B$ are conjugate over $\mathbb Q_3$ but their reductions to $\mathbb F_3$ are not conjugate. Moreover, both have finite order (their order is $9$).
To see that their reductions mod 3 are not conjugate just compute the rank of 
$A-\textrm{id}$ and $B-\textrm{id}$ in $\mathbb F_3^{10\times 10}$. The rank is 8 in the first case and $7$ in the second, so they cannot be conjugate.
To see that they are conjugate over $\mathbb Q_3$ just compute minimal and characteristic polynomial. The minimal polynomial is $x^9-1$ in both cases, which implies that the matrices are semisimple. Therefore they are conjugate if and only if their characteristic polynomials coincide. But the characteristic polynomial is $(x-1)(x^9-1)$ in both cases.