Let $q$ be prime power, let $t \in \mathbb{N}$, and let $\omega \in \mathbb{F}_{q^t}*$. Consider the following invertible, $\mathbb{F}_q$-linear mapping on $\mathbb{F}_{q^t}$.
$g:\mathbb{F}_{q^t} \rightarrow \mathbb{F}_{q^t}$, $x \mapsto x^q\omega$
(The mapping $g$ is also $\mathbb{F}_{q^t}$-semilinear.) I believe (and would like to prove) that $g$ is cyclic as an element of ${\rm GL}_q(\mathbb{F}_{q^t})$, in the sense that there exists $v \in \mathbb{F}_{q^t}^*$ such that the elements
$v, vg, vg^2, \dots, vg^{t-1}$
form an $\mathbb{F}_q$-basis of $\mathbb{F}_{q^t}$. This is equivalent to saying that the minimal and the characteristic polynomial of $g$ coincide. Unfortunately, I have not yet found a proof (or a counterexample). I would appreciate any help/suggestions.
(Note that $g^t$ is given by $x \mapsto x \alpha$, where $\alpha = N_{\mathbb{F}_{q^t}:\mathbb{F}_q}(\omega)$ is the norm of $\omega$ over $\mathbb{F}_q$. Since $\alpha \in \mathbb{F}_q$, the minimal polynomial of $g^t$ is $x- \alpha$. Thus the minimal polynomial of $g$ divides $x^t-\alpha$. I would like to show that the minimal polynomial of $g$ is equal to $x^t-\alpha$.)
Thanks in advance!
If the minimal polynomial of $g$, call it $m(T)=\sum_{i=0}^dm_iT^i$, had degree $d<t$ then every element of $\Bbb{F}_{q^t}$ would be a zero of the polynomial $$ \tilde{m}(x)=\sum_{i=0}^d m_ig^{[i]}(x), $$ where $g^{[i]}(x)$ stands for the $i$-fold composition $g\circ g\circ \cdots \circ g$. But, $\tilde{m}(x)$ has degree $q^d$, so cannot have $q^t$ zeros unless $d\ge t$.
This also settles your main claim. I got the impression that you figured that part out. It's similar to a standard proof of existence of a normal basis for the extension $\Bbb{F}_{q^t}/\Bbb{F}_q$, so I won't waste bandwidth by including those details.