Let $E \supset F \supset G$ and $G = E^{\rm{Aut}(E/G)}$. Then is it true that $F = E^{\rm{Aut}(E/F)}$?

Question

Let $E/F$ be a field extension. Let us say $E/F$ is semi-Galois if $F$ is the fixed field of $\rm{Aut}(E/F)$.

Now consider a field tower $E/F/G$. If $E/G$ is semi-Galois, then is $E/F$ necessarily semi-Galois? What would be a counterexample if this is false?

Answer 1

Let $F$ be an algebraic closure of $\mathbb{Z}/2\mathbb{Z}$. Then,$F(X)/F$ is semi-Galois while $F(X)/F(X^2)$ is not a semi-Galois.