$F$ field, $K/F$ finite Galois extension and $G=\mathrm{Gal}(K/F)$. Let $H \leq G$ and $L=\mathrm{Fix}(H)$ (the fixed subfield of $K$ by elements of $H$).
I want to prove the following: $$\mathrm{Aut}(L/F)\cong N_G (H)/H.$$ I think the "Fundamental Theorem of Galois Theory" will be helpful, but I have no idea the way to prove there exist such isomorphism, or maybe it is enought to show that these groups have the same order. Thanks for any help
We start by constructing a group homomorphism from $N_G(H)$ to $Aut(L/F)$.
First a general remark: if $\sigma$ is any element of $G$, then the field $\sigma(L)$ is exactly the subfield of $K$ fixed by the subgroup $\sigma H\sigma^{-1}$ of $G$. This is obvious to check.
In particular, if $\sigma$ is in $N_G(H)$, then we have $\sigma H\sigma^{-1} = H$, and hence $\sigma(L)$ is equal to $L$, i.e. the restriction $\sigma\vert_L$ is an automorphism of $L/F$. Conversely, it is also clear by Galois theory that if $\sigma\vert_L$ is an automorphism of $L$, then $\sigma$ normalizes $H$.
Therefore we may define $\phi: N_G(H) \rightarrow Aut(L/F)$ sending any $\sigma$ to $\sigma\vert_L$. It is routine to check that $\phi$ is a well-defined group homomorphism. Also, it is surjective, because the extension $K/F$ being normal, every automorphism of $L/F$ lifts to an automorphism of $K/F$, which is an element of $N_G(H)$.
It only remains to identify the kernel of $\phi$. This is again easy to check: an element $\sigma \in G$ is in the kernel of $\phi$ if and only if $\sigma\vert_L$ is the identity map, i.e. $\sigma(l) = l$ for all $l \in L$. By Galois theory, the last condition is equivalent to saying that $\sigma \in H$.