Prove or disprove:
Let $A, B \in \operatorname{M}(3,\mathbb{Z})$ and $A \sim B$. $A$ has a square root in $\operatorname{M}(3,\mathbb{Z})$ iff $B$ has a square root in $\operatorname{M}(3,\mathbb{Z})$.
If $AX = XB$ for some invertible $X$ in $\operatorname{M}(3,\mathbb{Z})$ with $\det(X) = 1$, then it is true. What if $\det(X)$ is not equal to $1$ or $-1$ ?