Isomorphism between automorphism subgroup and quotient subgroup $F$ 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
 A: 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 $\sigma \in H$.
