Let $A$ and $B$ be two $n \times n$ matrices with entries in $\mathbb Z_p$, the $p$adic integers. Is it true that $A$ and $B$ are conjugate iff they're conjugate over $\mathbb Q_p$ and over $\mathbb F_p$?
A counterexample is $$ A=\left[\begin{array}{cc}0&2\\8&0\end{array}\right],\hspace{5mm}B=\left[\begin{array}{cc}0&4\\4&0\end{array}\right]\in M_2(\mathbb{Z}_2). $$ The matrices are conjugate in $\mathbb{Q}_2$ because they have the same eigenvalues $\pm 4$, and they are conjugate in $\mathbb{F}_2$ because both are $0$. But they are not conjugate in $\mathbb{Z}_2$ because $A\not\in 4\cdot M_2(\mathbb{Z}_2)$ but $B\in 4\cdot M_2(\mathbb{Z}_2)$.

$\begingroup$ Awesome, thanks. What if we say conjugate in $\mathbb Z/p^n \mathbb Z$ for all $n$? $\endgroup$ – Nick Addington Oct 11 '17 at 20:32

$\begingroup$ Or actually, in my application all the eigenvalues have (integer) length that's prime to p. Any thoughts there? $\endgroup$ – Nick Addington Oct 11 '17 at 20:34

9$\begingroup$ Conjugate in $\mathbb{Z}/p^n\mathbb{Z}$ for all $n$ should give conjugate in $\mathbb{Z}_p$ because we can choose a compatible family of conjugating matrices. The matrices $A=[1,2;0,1]$ and $B=[1,4;0,1]$ are conjugate in $\mathbb{Q}_2$, are equal in $\mathbb{F}_2$, and have integer eigenvalues coprimes to $2$, yet they are not conjugate in $\mathbb{Z}_2$. $\endgroup$ – Julian Rosen Oct 11 '17 at 20:47