If $E/F$ is a field extension, and $H$ is a subgroup of $\mathrm{Aut}(E/F)$, it is quite trivial to see that $H\subset \mathrm{Aut}(E/E^H)$.
Since the theorem only shows the inclusion relationship, I think there must be a lot of examples where $\subset$ is actually $\subsetneq$. But due to lack of knowledge I can't come up with one.
Could you help me with this? Best regards.
If $H$ is finite, we always have equality. See these notes by Noam Elkies.
There are counterexamples, when $H$ is infinite.
If we allow a transcendental extension, then the following example is easy to grasp. Let $E=\Bbb{Q}(x)$ be the field of fractions of the polynomial ring $\Bbb{Q}[x]$. Let $\sigma$ be the automorphism gotten by extending $x\mapsto x+1$ in the obvious way, and let $H=\langle\sigma\rangle$ be the infinite cyclic group. Then it is not difficult to show (for an argument see an earlier answer of mine) that $E^H=\Bbb{Q}$. But $E$ has many other $\Bbb{Q}$-automorphisms. They are all of the form $$ x\mapsto \frac{ax+b}{cx+d} $$ with $ad-bc\neq0$ (= a quotient group of invertible 2x2 matrices by scalar matrices).
If you want an example of an algebraic extension, then it is a bit more complicated. Let $E$ be the algebraic closure of $\Bbb{F}_p$, and let $H$ be the group of automorphisms generated by the Frobenius mapping $F:z\mapsto z^p$. Because a degree $p$ polynomial can have at most $p$ fixed points we see that $E^H=\Bbb{F}_p$. But $H$ is not all of $Aut(E/\Bbb{F}_p)$. The recipe for all $\Bbb{F}_p$-automorphisms of $E$ is described in this answer by Ted. Here $H$ is only a dense subgroup of $Aut(E/\Bbb{F}_p)$ (w.r.t. the Krull topology). You need to read a bit about Galois theory of infinite algebraic extensions for the use of topological concepts to make sense here.
