$\DeclareMathOperator\GL{GL}$Let $A\in \GL\_d(\mathbb{Z})$ have finite order $n.$ Suppose that $k\in \mathbb{Z}$ is relatively prime to $n.$ Is $A^k$ conjugate to $A$ in $\GL\_d(\mathbb{Z})$?


For $d\leq 4$ the answer is yes. Indeed the papers *"On the finite subgroups of $\GL(3,\mathbb{Z})$"* by K. Tahara, 1971 and *"Conjugacy Classes of Torsion in $\GL\_n(\mathbb{Z})$"*, by Q. Yang, 2015 list all torsion elements up to conjugacy for the cases $d=2,3$ and $d=4$ respectively. There are not many cases that need to be checked, so I checked each case and the answer turns out to be "yes" in all of these cases. But I have no general argument for why things work out when $d\leq 4$, only computations.



Finite order elements of $\mathrm{GL}_d(\mathbb{Z})$ that are conjugate to powers of themselves
The answer is "no" in general. There may be an elementary way of seeing this, but I will frame this in representation theoretic terms and will describe a general construction.


The question is equivalent to the following question: let $C\_n$ be a cyclic group of order $n$, let $g$ be a generator, and let $\rho\colon C\_n\to {\rm GL}\_d(\mathbb{Z})$ be an integral representation of $C\_n$, defined by $g\mapsto A$. Let $\sigma\in {\rm Aut}(C\_n)$ be defined by $g\mapsto g^k$. Then is the integral representation $\rho\circ\sigma$ of $C\_n$ isomorphic to $\rho$? Let me now explain a general construction of representations for which the answer is "no".


Set $d=\phi(n)$, the Euler phi function. For example if $n$ is prime, then we will have $d=n-1$. Fix a primitive $n$-th root of unity $\zeta\_n$, and let $I$ be an ideal in $\mathbb{Z}[\zeta\_n]$, the ring of integers of the $n$-th cyclotomic field $\mathbb{Q}(\zeta\_n)$. Let $C\_n$ act on $I$ by letting $g$ act by multiplication by $\zeta\_n$. Since $I$ is an ideal, multiplication by $\zeta\_n$ preserves it as a set, so this defines a representation $\rho\_I\colon C\_n\to {\rm GL}\_d(\mathbb{Z})$. Moreover, it is not hard to see that if $I$ and $J$ are two such ideals, then $\rho\_I$ is isomorphic to $\rho\_J$ if and only if $I$ and $J$ represent the same class in the ideal class group of $\mathbb{Q}(\zeta\_n)$. Finally, given $k$ coprime to $n$, there exists an element $\sigma$ of the Galois group of $\mathbb{Q}(\zeta\_n)$ such that $\sigma\zeta\_n = \zeta\_n^k$. This $\sigma$ also defines an automorphism of $C\_n$ via $g\mapsto g^{k}$. It is easy to see that $\rho\_I\circ \sigma=\rho\_{\sigma^{-1}I}$. Therefore, to produce an example of the type you desire, you can take a cyclotomic field in which the Galois group acts non-trivially on the class group and let $I$ be a representative of any class in the class group that is not preserved by the Galois group.


The smallest example that you can produce this way has $n=23$, $d=22$. Indeed, the cyclotomic field $\mathbb{Q}(\zeta\_{23})$ is the first cyclotomic field with non-trivial class group, which has order $3$, and complex conjugation acts by $-1$ on it. So take any non-principal ideal $I$ in $\mathbb{Z}[\zeta\_{23}]$, write down a $\mathbb{Z}$-basis for it, and write out the matrix $A$ of multiplication by $\zeta\_{23}$ on this basis. Then $A$ will not be conjugate to $A^{-1}=A^{22}$ in ${\rm GL}\_{22}(\mathbb{Z})$.


