We have this theorem.
Let $L|K$ a field extension with $[L:K]<\infty$ and $G=\text{Aut}(L|K)$. We let $G$ act on $L$. Then there is a trivial stabilizer.
The proof is the following, I would like to get help with the vector space argument (the $\neq$ part), I don't see how you can argue like that without knowing beforehand that the automorphism group is finite (in fact, this result is used to show that it's finite in my lecture notes later).
If $G=\{\text{id}\}$ we are done, so assume $G\neq\{\text{id}\}$ and $L\neq K$. For $f\in G$ let $L_f$ be the set of all $y\in L$ that are fixed under $f$. Obviously, for all $f$ we have that $L_f \leq L$ is a field. Moreover, if $f\neq \text{id}$, $L_f$ will be a proper subfield of $L$. We can consider the $L_f$ to be subspaces of the $K$-vector space $L$. Thus, $$L\neq \bigcup_{f\in G\setminus\{\text{id}\}}L_f,$$ so that there exists a $y\in L$ that lies in no nontrivial $L_f$ and thus, the stabilizer of $y$ is trivial.