Does there exist a field $k$ such that there is finitely many (up to equivalence of field extensions) finite field extensions of prime degree of $k$ that are isomorphic to $k$ as fields (and at least one such extension exists)? I can think of examples with infinitely many such extensions (e.g. $\mathbb{C}((t))$ to which we can adjoin various roots of $t$).
We require the degree to be prime because given an extension of degree $d$ as in the question, it can be applied to itself inductively so we get extensions of degree $d^i$ for all positive integers $i$.
This follows from the Puiseux series theorem, that $\bigcup_n \Bbb{C}((X^{1/n}))$ is algebraically closed.
In term of the theory of complete local fields : for a finite extension of $T(a)/T$ of degree $q$, the minimal polynomial of $a$ is in some $K_n=\Bbb{C}((X^{1/n}))$ so that $K_n(a)/K$ is a totally ramified extension of degree $q$, thus $\pi_{K_n(a)}^q = X^{1/n} u$ for some $u\in O_{K_n(a)}^\times$ ie. $u=z(1+\pi_{K_n(a)} b)$ with $z\in \Bbb{C}^*,a\in O_{K_n(a)}$, since $O_{K_n(a)}$ is a complete DVR then $u^{1/q} =z^{1/q} \sum_{m\ge 0}{1/q\choose m} \pi_{K_n(a)}^m b^m\in O_{K_n(a)}^\times$ thus $\varpi_{K_n(a)}=\pi_{K_n(a)} u^{-1/q}\in O_{K_n(a)}$ is an uniformizer which satisfies $\varpi_{K_n(a)}^q=X^{1/n}$ and hence $K_n(a)=K_n(\varpi_{K_n(a)})=K_n(X^{1/(nq)})$, $T(a)=T(X^{1/(nq)})$ which implies that $q=p^r$ and $T(a)=T(X^{1/p^r})$.