Let $F$ be a finite field and let $G=GL_n(F)$ be the general linear group over $F$. There is a group action $G\times F^n \to F^n$ given by $(T,x) \mapsto Tx$. I suspect that this action is not transitive, but am unsure if my proof works.
Lemma: Let $G$ be a finite group that acts transitively on a set $X$. Then $G = \cup_{x\in X} \text{Stab}(x)$ if and only if $|X| = 1$. ($\text{Stab}(x) = \{ g \in G \mid gx = x\}$).
This lemma can be proved using Burnside's lemma. In my case, with $G = GL_n(F)$ and $X = F^n$, we have $\text{Stab}(0) = G$ and therefore $G = \cup_{x\in X} \text{Stab}(x)$. Since $|F^n| >1$, this action cannot be transitive by the above lemma.
Is there an easier way to show that this action is not transitive?
Suppose $v\in F^n$ is nonzero. Does there exist a $g\in \mathrm{GL}_n(F)$ such that $g(0)=v$? So then what must the orbit of $0$ be?
Indeed, there are two orbits, $\{0\}$ and $\{$everything else$\}$. Try proving it.