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?
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 $F$).