Let $p$ be a prime number and $n\ge 1$ an integer. Call $N:=p^n-1$.
Can the cyclic group of order $N$ be contained in $\text{GL}_N(\mathbf{Z})$? Can it be contained in $\text{PGL}_N(\mathbf{Z})$?
What can be said if $\mathbf{Z}$ is replaced by the ring of integers of a number field? For example a cyclotomic field generated by a primitive $m$-th root of unity.
The question originated from the reading of a paper of Serre, building on Minkowski's results on finite subgroups of $\text{GL}_N(\mathbf{Q})$. Namely, I was trying to see if the condition that the cyclic group of order one less than a prime power ($N=p^n-1$ as in the body of the question) is contained in $\text{PGL}_N(\mathbf{Z})$ or $\text{GL}_N(\mathbf{Z})$, puts any constraints on $p$ or on $n$. For instance, whether if $p$ is large, $n$ is forced to be one.
Consider $$\begin{pmatrix} 0 & 1 & 0 & \dots \\ 0 & 0 & 1 & 0 & \dots \\ & & \dots & & \\ & & \dots & 0 & 1 \\ 1 & 0 & \dots & \end{pmatrix}$$ i.e. 1's on the superdiagonal and 1 in the bottom left corner. It corresponds to the permutation $(1\, 2\, 3 \,\dots \, N)$ of the basis vectors. If $N$ is odd, it even has determinant 1, otherwise determinant $-1$. And of course, it has order $N$, thus generating a subgroup isomorphic to $\Bbb{Z}/N\Bbb{Z}$. This naturally works for any $N$.
Any matrix contained in $GL_N(\Bbb{Z})$ is also "contained" in $PGL_N(\Bbb{Z})$, since the former surjects onto the latter.