Let $E$ be an extension field of $F$ and suppose that $K$ and $L$ are two intermediate fields. If there exists an element $\sigma \in G(E/F)$ such that $\sigma(K) = L$ then K and L are said to be conjugate fields. Prove that $K$ and $L$ are conjugate iff $G(E/K)$ and $G(E/L)$ are conjugate subgroups of $G(E/F)$.
This direction ($\Rightarrow$):
Consider $\sigma G(E/K) \sigma^{-1}$ where $\sigma \in G(E/F)$ and $\sigma(K) = L$ and consider some element $\alpha \in L$. I believe that if I can show that $\sigma G(E/K) \sigma^{-1}$ fixes $\alpha$, then we are done.
So consider any $\gamma \in G(E/K)$. As $\sigma (K) = L$ then $K = \sigma^{-1}(L)$.
Thus, $\sigma^{-1}(\alpha)=\beta$ for some $\beta\in K$.
And then, $\gamma(\sigma^{-1}(\alpha))=\gamma(\beta)=\beta$ as $\gamma$ fixes all $\beta \in K$
Finally, $\sigma(\gamma(\sigma^{-1}(\alpha)))=\sigma(\beta)=\alpha$. So, $\sigma G(E/K) \sigma^{-1}$ consists of all automorphisms of E that fix $\alpha \in L$ and so it is $G(E/L)$
This direction ($\Leftarrow$):
Well, if $G(E/K)$ and $G(E/L)$ are conjugate then there exists some $\sigma \in G(E/F)$ such that $\sigma G(E/K) \sigma^{-1} = G(E/L)$. So consider some $\alpha \in K$. If $\alpha$ is also in $F$ (as $K$ is an extension field of $F$ because it is an intermediate field between $E$ and $F$) then $\alpha$ is fixed by $\sigma G(E/K) \sigma^{-1}$ and as $\alpha \in F$ then $\alpha$ is also in $L$.
If $\alpha \in K\backslash F$ then I don't know how to proceed.
I believe that the you need the assumption that $F \subseteq E$ is a Galois extension. Indeed consider the non-Galois extension $\mathbb{Q} \subseteq E = \mathbb{Q}(\sqrt[3]{2})$. The Galois group is the trivial group of a single element. Hence $G(E/\mathbb{Q}) = G(E/E)$ and in particular they are conjugates. However it is not true that there exists $\sigma \in G(E/\mathbb{Q})$ s.t. $\sigma(E) = \mathbb{Q}$, as then $\sigma$ must be the identity and we know $E \not = \mathbb{Q}$
If we assume the extension is Galois the claim holds. In particular if $\sigma G(E/K)\sigma^{-1} = G(E/L)$, then we have that $\sigma(K) = L$.
First we'll prove $\sigma(K) \subseteq L$. Consider $\sigma(\alpha)$ for some $\alpha \in K$. Also let $\gamma$ be an arbitrary element of $G(E/L)$. Then we have that $\exists \beta \in G(E/K)$ s.t. $\gamma = \sigma \beta \sigma^{-1}$. Then we have:
$$\gamma(\sigma(\alpha)) = \sigma \beta \sigma^{-1}(\sigma(\alpha)) = \sigma(\alpha)$$
So we have that $\sigma(\alpha)$ is in the fixed field of $G(E/L)$, which is $L$ itself. (Here's where we use the fact that the extension is Galois).
Now similarly you can prove that $\sigma^{-1}(L) \subseteq K$ to finish off the proof.