Consider the Galois extension of $K/L$ with $G:=Gal(K/L)$. Let $L\subset E \subset K$. Define $H:=\{\sigma\in G: \sigma(\alpha)=\alpha, \forall \alpha\in E\}$ with $H<G$. Let $C:=Gal(K/E)$. I am working on how to prove that $$ N_G(C)=\{g\in G: gCg^{-1}=C\} $$ is same as subgroup $H$.
What I am confused about currently is that using this theorem, we will get two different one-to-one correspondences by Fundamental Theorem of Galois Theory: $$ \{L \subset E \subset K\} \longleftrightarrow \{G > C>1\} $$ and $$ \{L \subset K^H \subset K\} \longleftrightarrow \{G > H>1\} $$
To prove $H=N_G(C)$, I am stuck on both inclusion direction.
(1) For $h\in H$ and $c\in C$, we have for every $\alpha \in E$, $$ hch^{-1}(\alpha)=hc(\alpha) $$ but how to show that is just $c(\alpha)$?
To finish your argument for (1), since $c \in C = \operatorname{Gal}(K/E)$ and $\alpha \in E$, we have $c(\alpha) = \alpha$ so $$hch^{-1}(\alpha) = \alpha = c(\alpha).$$ Hence, $H \leq N_G(C)$.
Can you figure out the second inclusion? Probably not, because here you have $H = C$ which is not necessarily equal to $N_G(C)$.
For the correct result, redefine $H = \{\sigma \in G \mid \sigma(\alpha) \in E,\, \forall \alpha \in E\}$. Now, we will show $H = N_G(C)$.
Let $h \in H$, $c \in C$ and $\alpha \in E$. Then, $h^{-1}(\alpha) \in E$ because $h^{-1} \in H$. So, $ch^{-1}(\alpha) = h^{-1}(\alpha)$ because $c$ fixes $E$ and therefore, $hch^{-1}(\alpha) = hh^{-1}(\alpha) = \alpha$. Thus, $H \leq N_G(C)$.
Now, let $n \in N_G(C)$, $c \in C$ and $\alpha \in E$. Then $n^{-1} \in N_G(C)$ so, $n^{-1}cn(\alpha) = \alpha$. Applying $n$ on both sides, we get $c(n(\alpha)) = n(\alpha)$, meaning $c$ fixes $n(\alpha)$. Therefore, $n(\alpha) \in K^C = E$. Thus, $N_G(C) \leq H$. Hence, $H = N_G(C)$.