= Solution
A <K-isomorphism> $L\to L'$ is a <field isomorphism> fixing every element of $K$. The <field automorphism group> $\operatorname{Aut}_K(L)$ 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 $s\leq d$. This action is faithful because fixing $\alpha$ fixes the generated field. Thus $\operatorname{Aut}_K(K(\alpha))$ 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 $p$, but
$$
K(\alpha)=\mathbb F_{p^p}(t),\qquad K(\beta)=\mathbb F_p(\beta).
$$
The 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. \b[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>. \b[There is no finite-base-field example of this kind.]
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