Let $F ⊂ L$ be Galois and $F ⊂ K ⊂ L$ be an intermediate field. Given $σ_1K, · · · , σ_rK$ are the distinct conjugates of $K$ where $σ_1 = e$ and $σ_j ∈ Gal(L|F), 1 ≤ j ≤ r$, I wish to prove $r = [Gal(L|F) : N]$ where $N$ is the normaliser subgroup $N_{Gal(L|F )}(Gal(L|K))$.
First, I want to show $Gal(L|F)$ acts on the set of conjugates {$ σ_1K, · · · , σ_rK $}. Second, if we can show the isotropy subgroup of $K$ or $Stab_{Gal(L|F )}(K)$ is the normalizer subgroup $N$, then I think the desired result can be deduced from the Fundamental Theorem of Group Actions, if I am not mistaken. But I am not sure of the details hmm~
Let $G=Gal(L/F)$ and $H=Gal(L/K)$ be the subgroup corresponding to the intermediate field $K$. Let $\sigma\in G$ be arbitrary. By basic properties of group actions (also easy to verify as an exercise, ask if you need help in this step) $\sigma(K)$ is the field of fixed points of the conjugate subgroup $\sigma H\sigma^{-1}$.
The Galois correspondence is bijective, so for any two automorphisms $\sigma,\sigma'\in G$ we have $\sigma(K)=\sigma'(K)$ if and only if $\sigma H\sigma^{-1}=\sigma' H\sigma'^{-1}$ if and only if $\sigma'\sigma^{-1}$ belongs to the normalizer $N_G(H)$. This means that the conjugate fields $\sigma(K)$ are in a bijective correspondence with the (left) cosets $\sigma N_G(H)$ of the normalizer.
Therefore the number of conjugate intermdiate fields $\sigma(K)$ is equal to $r=[G:N_G(H)]$ as claimed.