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A K-isomorphism $L\to L^{\prime }$ is a field isomorphism fixing every element of $K$. The field automorphism group ${\mathrm{Aut}}_K(L)$ consists of such isomorphisms $L\to L$, with composition.

If an isomorphism sends $\alpha$ to $\beta$, it takes every polynomial relation over $K$ for $\alpha$ to the same relation for $\beta$; the monic minimal polynomials coincide. Conversely, if their common minimal polynomial is $m$, evaluation induces field isomorphisms $K[X]/(m)\cong K(\alpha )$ and $K[X]/(m)\cong K(\beta )$. Their composite sends $\alpha$ to $\beta$ and fixes $K$.

An automorphism of $K(\alpha )$ sends $\alpha$ to a root of $m$ inside $K(\alpha )$ and is determined by this image. More precisely the field automorphism group permutes the set of roots of $m$ lying in this field, a set of size $s\le d$. This action is faithful because fixing $\alpha$ fixes the generated field. Thus ${\mathrm{Aut}}_K(K(\alpha ))$ embeds in $S_s$, and then in $S_d$ by fixing $d-s$ extra labels. In particular it is finite. This argument also covers inseparable minimal polynomials.

For an example not isomorphic even as abstract fields, take $K={\mathbb{F}}_p(t)$, let $\alpha$ generate ${\mathbb{F}}_{p^p}$ over ${\mathbb{F}}_p$, and put $\beta =t^{1/p}$. The constant-field minimal polynomial of $\alpha$ remains irreducible over the rational function field: if it factored over ${\mathbb{F}}_p(t)$, each monic factor's coefficients would be algebraic over ${\mathbb{F}}_p$, being symmetric functions of constant algebraic roots. A rational function algebraic over ${\mathbb{F}}_p$ is constant, so those coefficients would lie in ${\mathbb{F}}_p$, contradicting the original irreducibility. The polynomial $X^p-t$ is irreducible by Eisenstein criterion at $t$ in ${\mathbb{F}}_p[t]$. Both degrees are $p$, butThe first field has $p^p$ roots of $X^{p^p}-X$; the second has only $p$. Indeed a rational function algebraic over a finite constant field must be constant: a nonconstant rational function makes the transcendental variable algebraic over its own rational-function field, and therefore cannot itself be algebraic over the constants. The root counts are isomorphism invariants. These extensions are not isomorphic.

If $K={\mathbb{F}}_q$ is finite, every degree-$d$ extension is a field with $q^d$ elements. Such fields are splitting fields of $X^{q^d}-X$ over $K$, so uniqueness of the splitting field gives a K-isomorphism. There is no finite-base-field example of this kind.