Let $q$ be a prime power. Let $\mathbb{F}\_q$ be the finite field with $q$ elements. Then $\mathbb{F}\_{q^n}$ is a field extension of $\mathbb{F}\_q$ of degree $n$ and can be considered as an $n$-dimensional vector space $V$ over $\mathbb{F}\_q$. Now consider the action of $GL(V)$ on $V$. Any element $x$ in $\mathbb{F}\_{q^n}$ acts on itself by multiplication and hence defines an element in $GL(V)$; we denote this element as $g\_x$. Suppose now an element $h\in GL(V)$ commutes with $g\_x$ for some $x\neq0,1\in \mathbb{F}\_{q^n}$. Does it imply that $h=g\_y$ for some $y\in \mathbb{F}\_{q^n}$ as well? Thanks. 



General linear group action on extensions of finite fields
Let $F$ be any field and $F<E$ a finite field extension.
Fix $x\in E^\*$ and consider the multiplication operator $g\_x\in \text{GL}\_F(E)$,
the group of invertible $F$-linear transformations of $E$.
The centralizer of $g\_x$ could be naturally identified with the subgroup $\text{GL}\_{F[x]}(E)<\text{GL}\_F(E)$, where $F[x]<E$ is the subfield of $E$ generated by $x$ over $F$. We thus have
$$ E^\* \simeq \text{GL}\_E(E)<\text{GL}\_{F[x]}(E)<\text{GL}\_F(E).$$
The inclusion $\text{GL}\_E(E)<\text{GL}\_{F[x]}(E)$ is an equation iff $F[x]=E$.
To see the "only if" part it is enough to recall that $\text{GL}\_n$ is never commutative for $n\geq 2$.
Thus every element in the centralizer of $g\_x$ is of the form $g\_y$ for $y\in E^\*$ iff $E=F[x]$, that is $x$ generates $E$ over $F$.


Specializing to $F=\mathbb{F}\_q$ and $E=\mathbb{F}\_{q^n}$,
we get that every element in the centralizer of $g\_x$ is of the form $g\_y$ for $y\in \mathbb{F}\_{q^n}^\*$ iff $x$ generates $\mathbb{F}\_{q^n}$ over $\mathbb{F}\_q$.


