Let $K/F$ be a field extension, where $|F|=p$ and $|K|=q=p^r$.
Let $H$ be a subgroup of $\operatorname{Aut}(K/F)$. Let $K^H=\{k\in K: \forall h \in H[h(k)=k\}$. We see that $K^H$ is a subfield of $K$.
I want to prove that $H=\operatorname{Aut}(K/K^H)$ WITHOUT using the theorems of Galois Theory. However, we can take advantage of the properties of finite fields.
Attempt
I have shown that $H \subseteq\operatorname{Aut}(K/K^H)$. Since these sets are finite, to finish the proof it is enough to show $|H|=|\operatorname{Aut}(K/K^H)|$. By using Frobenius map, one can see that $|\operatorname{Aut}(K/K^H)|=[K:K^H]$.
So to finish proof, it is enough to show $|H|=[K:K^H]$. I would ordinarily just cite the fixed field theorem here but, alas, it seems to be a theorem of Galois Theory. Is there an easy way around this stumblingblock, taking advantage of finite fields?
Let $G = \operatorname{Aut}(K/K^H)$ and write $|K^H| = p^s$. Then $|G| = [K:K^H] = r/s$.
The group $G$ is cyclically generated by the Frobenius map $\varphi : x \mapsto x^{p^s}$. From this we see that $H$ is the unique subgroup of order $e = |H|$ in $G$. Consequently, $H$ is generated by $\varphi^{r/(se)}$ such that $$ K^H \:=\: \{ x \in K : \varphi^{r/(se)}(x) = x \} \:=\: \{ x \in K : x^{p^{r/e}} = x \} \,. $$ This means that $K^H$ is the unique subfield of $K$ with $p^{r/e}$ elements. Hence, $[K:K^H]=\frac{r}{r/e} = |H|$.