Let F/K be a finite dimensional Galois extension and $E$ be an intermediate field. Prove that there is a unique smallest field $L$ such that $E \rightarrow L \rightarrow F $ and $L/K$ is Galois. Moreover, $Aut(F/L) = \bigcap_{\sigma\in AutF/K } \sigma Aut(F/E)\sigma^{-1}$.
For the first question I guess the answer is just to take the normal closure of $E/K$ however for the second I have no idea.
Any help will be much appreciated thanks
If you already know the main theorem of Galois theory, you know that the Galois subextensions $$K\hookrightarrow E \hookrightarrow F$$ correspond to the subgroups $\operatorname{Gal}(F/E)$ of $\operatorname{Gal}(F/K)$, and the Galois subextensions $$K\hookrightarrow L \hookrightarrow F$$ to the normal subgroups $\operatorname{Gal}(F/L)$. So what do you want is the maximal normal subgroup contained in $\operatorname{Gal}(F/E)$.
Now the problem is transformed to a problem on (finite) groups: Given a group $G$ and a subgroup $H$, the maximal normal subgroup contained in $H$ is $$N:=\bigcap_{\sigma\in G} \sigma H\sigma^{-1},$$ which it is straightforward from the definition of normal subgroup.