General linear group action on extensions of finite fields 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. 
 A: 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 $\mathbb{F}_q$.
A: Let $k=\Bbb{F}_{q},K=\Bbb{F}_{q^n}$, $x\in K$ and $g_x\in M_n(k)\cong End_k(K)$ the matrix of the multiplication by $x$.
If $k(x)$ is smaller than $K$ then every $k(x)$-linear endomorphism of $K$ commutes with $x$, most of them are not of the form $g_x$.

Assume that $K=k(x)$ 

$g_x$ has an eigenvector $v\in K^n$ with eigenvalue $x$. Since $g_x\in M_n(k)$ then $$g_x(v^{q^l})  =(g_x(v))^{q^l}=(xv)^{q^l}=x^{q^l}v^{q^l}$$
Since the $x^{q^l}$ are distinct the $v^{q^l}$ must be distinct and we have diagonalized $g_x$ $$g_x(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l x^{q^l} v^{q^l}$$
If $h\in M_n(k)$ commutes with $g_x$ then $h$ has an eigenvector which is an eigenvector of $g_x$,  wlog we can assume it is $v$ so that $h(v)=av$ for some $a \in F$ where $F$ is the splitting field of $h$'s minimal polynomial. As before we obtain the diagonalization from $$h(v)=av \implies h(v^{q^l}) = a^{q^l} v^{q^l}, \qquad h(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l a^{q^l} v^{q^l}$$
$v^{q^n}= v$ implies that $a^{q^n}=a$ thus $a\in K$. Whence $a = f(x)$ for some $f\in k[T]$ and $$h = f(g_x)=g_a$$
