 Solution

A field homomorphism fixing $K$ is injective, since its kernel is an ideal of a field and cannot contain one. It is also $K$-linear. Equal finite field extension degrees then make it a bijection, hence a $K$-isomorphism.

Equal degrees alone do not ensure that any such homomorphism exists. For example $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ both have degree two, but a $\mathbb{Q}$-isomorphism would require a square root of two in $\mathbb{Q}(\sqrt{3})$. Squaring $a+b\sqrt{3}$ with rational $a,b$ would give $2ab=0$ and $a^2+3b^2=2$. Neither $a^2=2$ nor $b^2=2/3$ has a rational solution. Thus the two degree-two extensions are not isomorphic over $\mathbb{Q}$.