Suppose $E_1, E_2 \subset E$ are proper subfields. In general, if one has an isomorphism $\sigma:E_1\to E_2$, is it possible to extend it to an isomorphism $\psi:E\to E$ s.t. $\psi|_{E_1} = \sigma$ ?
Ideally, is it possible to get such a $\psi$ that also does not "affect" elements in $E\backslash E_1$ ? I thought $$ \psi(e):=\begin{cases} e, & e \in E\backslash E_1, \\ \sigma(e), & e \in E_1,\end{cases} $$ might work but unfortunately it looks like it does not in general.
Yes, if $E$ is an algebraic and normal extension of the fixed subfield of the morphism $E_1\to E_2$, but not in general.
Here is an example when you can't extend. Consider $E = \mathbb{Q}( \sqrt{1+ \sqrt{2}})\ $, $E_1 = E_2 = \mathbb{Q}(\sqrt{2}) $ and $E_1 \to E_2$, $\sqrt{2} \mapsto - \sqrt{2}$.
Let's show the existence of an extension. There are two fundamental facts about algebraic extensions of fields.
If $E/E_1$ is an algebraic extension then any $E_1 \to F$ is a morphism to an algebraically closed field $F$ can be extended to a morphism from $E$ to $F$.
Let $E/E'$ an algebraic normal extension and $F\supset E$ an extension of $E$. Let $\phi \colon E \to F$ a morphism of fields such that $\phi_{|E'} = Id_{E'}$. Then $\phi(E) = E$.
We are ready for the proof: Consider an extension of the morphism $E_1\to E_2$ from $E$ to an algebraic closure of $E$. Since the extesion $E/E'$ is algebraic and normal image of $E$ will conincide with $E$. Done.