= 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 $\mathbb Q(\sqrt2)$ and $\mathbb Q(\sqrt3)$ both have degree two, but a $\mathbb Q$-isomorphism would require a square root of two in $\mathbb Q(\sqrt3)$. Squaring $a+b\sqrt3$ 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 \b[the two degree-two extensions are not isomorphic over $\mathbb Q$].
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