Any two Singer cyclic subgroups of GL(n,q) are conjugate Cyclic subgroups of $\operatorname{GL}(n,q)$ of order $q^n - 1$ are called Singer cyclic subgroups. The following statement seems to be well-known:

Any two Singer cyclic subgroups of $\operatorname{GL}(n,q)$ are conjugate.

I don't know how to prove this in general. Sometimes, $q^n - 1$ is prime such that Sylow can be used (Example: $q=2, n=3$).
Some more background:


*

*Elements of order $q^n - 1$ (and thus Singer cyclic subgroups) always exist: Let $\alpha$ be a primitive element of $\operatorname{GF}(q^n)$ and look at the $\operatorname{GF}(q)$-linear map $\operatorname{GF}(q^n) \to \operatorname{GF}(q^n)$ given by $x \mapsto \alpha x$.

*$q^n - 1$ is the largest element order in $\operatorname{GL}(n,q)$, as by Cayley-Hamilton, the $\operatorname{GF}(q)$ vector space spanned by $I,A,A^2,A^3,\ldots$ has at most dimension $n$.

*The elements of order $q^n - 1$ are not necessarily conjugate. For example, in $\operatorname{GL}(3,2)$ there are two conjugacy classes of elements of order $7$. However, the generated Singer cyclic subgroups are conjugate.
 A: Here is a slightly expanded version of the proof mentioned by Dietrick Burde using the representation theory of cyclic groups.
A cyclic subgroup of order $q^n-1$ clearly clearly acts irreducibly on ${\mathbb F}_q^n$, so the minimal polynomial of a generator $g$ is a factor of $x^{q^n-1}-1$ of degree $n$ that is irreducible over ${\mathbb F}_q$.
So the subalgebra of $A_g \le M_n(q)$ generated by $g$ is isomorphic to the field of order $q^n$.
Since there is a unique such field up to isomorphism, for any other $g' \in {\rm GL}(n,q)$ of order $q^n-1$ generating a matrix algebra $A_{g'}$, there is an ${\mathbb F}$-isomorphism $\phi:A_g \to A_{g'}$, and the matrix defined by $\phi$ conjugates the set $\langle g \rangle = A_g \setminus \{ 0 \}$ to $\langle g' \rangle$.
A: The fact that all Singer "cycles" are conjugate follows from representation theory of cyclic groups. There is a detailed proof in the book "Finite groups" by B. Huppert, 1967, II. $§ 7$, page 187.
There is also a proof, that a Singer subgroup of $GL(n,q)$ is a maximal torus of the reductive group $GL(n,q)$, and that any two maximal tori are conjugate.
Reference: G. Hiss,  Finite groups of Lie type, section $1.2$.
