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$\DeclareMathOperator\GL{GL}$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.

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    $\begingroup$ Just as a motivation, if I'm correct, this is always true in $\mathrm{GL}_d(\mathbf{Q})$, due to the fact that the set of roots of every cyclotomic polynomial is stable under inversion. (And this is clearly false in $\mathrm{GL}_d(\mathbf{C})$ for every $d\ge 1$, just use a scalar matrix of finite order $\ge 3$.) $\endgroup$
    – YCor
    Feb 4 at 10:24
  • $\begingroup$ @YCor : What you say is true, but I think you need a little more than stable under inversion: the roots of irreducible cyclotomic polynomials are all Galois conjugate to each other (Galois groups being over $\mathbb{Q}$). $\endgroup$ Feb 4 at 10:29
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    $\begingroup$ @GeoffRobinson Oh, yes, I focussed on $A$ being conjugate to $A^{-1}$ after Alex B.'s answer. So, I should have said that the set of roots of $\Phi_n$ is stable under $x\mapsto x^m$ for $m$ coprime to $n$. $\endgroup$
    – YCor
    Feb 4 at 10:37
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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 ${\rm GL}_{22}(\mathbb{Z})$.

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    $\begingroup$ Thank you very much! $\endgroup$ Feb 4 at 0:27

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