Conjugate subgroups of Galois Group iff Isomorphic Extensions Problem statement: Conjugate Groups of Galois Group if and only if Isomorphic Extensions
I don't quite understand its answer. My confusion is:

*

*Given an field isomorphism, how do we convert it into an conjugate element?


*Similarly, given an conjugate element, how do we convert it into an isomorphism? We can't just restrict the domain right?
 A: This is based on my comments. Let us assume that $L/F$ is a finite Galois extension and that $$F\subseteq K_1\subseteq L, F\subseteq K_2\subseteq L$$ Let us assume that the subgroups $\text{Gal} (L/K_1),\text{Gal} (L/K_2)$ are conjugate to each other and hence we have an automorphism $\sigma\in \text{Gal} (L/F) $ such that $$\sigma\text{Gal} (L/K_1)\sigma^{-1}=\text{Gal} (L/K_2)$$ By a very standard theorem of Galois theory (check your textbook or prove yourself) the left hand side equals $\text {Gal} (L/\sigma(K_1)) $ and therefore $\sigma(K_1)=K_2$. Note that this also involves Galois correspondence which maps intermediate fields of $L/F$ as fixed fields of subgroups of $\text{Gal} (L/F)$.
For the second part let us assume that $K_1,K_2$ are isomorphic under the map $\phi$ and $\phi $ fixes $F$. Note that the Galois extension $L/F$ is finite and hence there are only a finite number of subfields between $L$ and $F$ and thus all the intermediate extensions are simple. Then we have a member $a\in K_1$ such that $K_1=F(a)$. Let $p(x) \in F[x] $ be the minimal polynomial of $a$ over $F$. And let $b=\phi(a) $. Then we can see that $$K_2=\phi(K_1)=\phi(F(a))=F(\phi(a))=F(b)$$ and $$p(b) =p(\phi(a)) =\phi(p(a)) =\phi(0)=0$$ so that $b$ is a conjugate of $a$. Since $L/F$ is Galois there is an automorphism $\sigma\in \text {Gal} (L/F) $ such that $\sigma(a) =b$ and therefore $\sigma(F(a)) =F(b) $ ie $\sigma(K_1)=K_2$ and thus $\phi$ is nothing but the restriction of $\sigma$ to $K_1$.
Now we have $$\text{Gal} (L/K_2)=\text{Gal} (L/\sigma(K_1))=\sigma\text{Gal} (L/K_1)\sigma^{-1}$$ and hence the subgroups $\text{Gal} (L/K_1),\text{Gal} (L/K_2)$ are conjugate.

Essentially you can see that conjugate subfields in a Galois extension are based on elements like $a, b$ which are conjugate to each other (ie they have same minimal polynomial over base field $F$).
