Let $F_1$, $F_2$ be subfields of $K$ and $\sigma\colon F_1\to F_2$ be a field isomorphism. Then I would intuitively guess that there should be a nice one-to-one correspondence between the following two sets:
Set 1. The set of all the automorphisms $\phi$ of $K$ such that the following diagram commutes:
Set 2. The set of all automorphisms $\psi$ of $K$ keeping $F_1$ fixed.
However, I am unsuccessful in my attempts to find any such correspondence (and getting frustrated now!). Is there any?
On
you cannot prove it, because it is not true. Take for example the nontrivial automorphism of $\mathbb{Q}(\sqrt{2})$ induced from $\sqrt{2} \mapsto -\sqrt{2}$. This cannot be extended to an automorphism of $\mathbb R$, so your first set is empty. However, the identity is an automorphism of $\mathbb R$ fixing $\mathbb{Q}(\sqrt{2})$, so your second set is nonempty.
If you can find a single such $\phi$, you have your correspondence: Add another column $F_1\hookrightarrow K$ to the left of your diagram, and attach it to your square via the identity map $F_1\to F_1$ and the map $\psi:K\to K$ being an element from your set 2. Observe that the composition of the entire bottom row $\phi\circ\psi:K\to K\to K$ is a corresponding element of set 1. It is not difficult to show that this correspondence is bijective.
As for the existence of $\phi$, well, there lies the problem. Consider, for instance, the case of rational functions over the reals: $$ K=F_1=\Bbb R(x)\\ F_2=\Bbb R(x^2) $$ Here $\sigma$ takes any rational function and simply doubles all the exponents of every $x$ that appears in it. For example, $$f=\frac{x^3-2}{x^4+x^2-x}\implies \sigma(f)=\frac{x^6-2}{x^8+x^4-x^2}$$ There can be no $\phi$ in this scenario, because by commutativity of the diagram and $K=F_1$ we would have $\phi=\sigma$, but this is not surjective. Your set 1 is therefore empty, while your set 2 contains the identity.