Let $L$ be an algebraically closed field and $\sigma \in \operatorname{Aut}(L)$. Let $F$ be the fixed field of $\sigma$, i.e., $F=\{x\in L:\sigma(x)=x\}$. Then I have to show that every finite extension $E/F$ is cyclic.
I proceed as following: To start with we may assume WLG that $E\subset L$, else take the copy of $E$ inside $L$ by the $F$-embedding of $\overline {F}=\overline {E}$ onto the algebraic closure of $F$ in $L$. First we show that $E/F$ is separable. For $\alpha \in E$ consider the set $\{\alpha,\sigma(\alpha),\sigma^2(\alpha),\ldots\}$, whose each element is a conjugate of $\alpha$ over $F$. Since there are only finitely many conjugates of $\alpha$ over $F$, there exists $n \in \mathbb {N} \cup \{0\}$ (choosing n smallest) such that $\sigma^n(\alpha)=\alpha$. Consider $f(x)=\prod\limits_{i=0}^{n-1}(x-\sigma^i(\alpha))\in F[x]$, and very clearly $m_{\alpha,F}(x)=f(x)$, which is a separable polynomial. Thus $E/F$ is separable. By primitive element theorem there exists $\beta \in E$ such that $E=F(\beta)$. WLG we may take $\beta=\alpha$ and $|E:F|=n$, so that our previous calculation reduce our work. Now consider the normal closure $N$ of $E/F$ inside $L$, i.e., $N=$Splitting field of $f(x)$ over $E$ inside $L$. Since $N$ is splitting field of a separable polynomial over $F$, $N/F$ is Galois. Identifying $\sigma$ with $\sigma \mid_{N}$ wish to show that $\operatorname{Gal}(N/F)$ is the cyclic group generated by $\sigma$. Clearly we've $\langle \sigma \rangle \leq \operatorname{Gal}(N/F)$. Conversely if $\tau \in \operatorname{Gal}(N/F)$ then $\tau(\alpha)=\sigma^i(\alpha)$ for some $i\in \{0,1,\ldots,n-1\}$. I stuck at this point.
How do I show that $\tau=\sigma^i$ to prove the claim?
Once we show that $N/F$ is cyclic it is clear that $E/F$ is also cyclic, moreover $N=E$ and we are done.
Exact point of difficulty: If $\tau(\alpha)=\sigma^i(\alpha)$, how can I show that $\tau(\sigma^j(\alpha))=\sigma^i(\sigma^j(\alpha))$ for all $j\in \{0,1,\ldots,n-1\}$?
Need some help. Thanks.
To start with we may assume WLG that $E\subset L$, else take the copy of $E$ inside $L$ by the $F$-embedding of $\overline {F}=\overline {E}$ onto the algebraic closure of $F$ in $L$. First we show that $E/F$ is separable. For $\alpha \in E$ consider the set $\{\alpha,\sigma(\alpha),\sigma^2(\alpha),\ldots\}$, whose each element is a conjugate of $\alpha$ over $F$. Since there are only finitely many conjugates of $\alpha$ over $F$, there exists $n \in \mathbb {N} \cup \{0\}$ (choosing n smallest) such that $\sigma^n(\alpha)=\alpha$. Consider $f(x)=\prod\limits_{i=0}^{n-1}(x-\sigma^i(\alpha))\in F[x]$, and very clearly $m_{\alpha,F}(x)=f(x)$, which is a separable polynomial. Thus $E/F$ is separable. Now consider the normal closure $N$ of $E/F$ inside $L$. Since $N$ is a splitting field of a separable polynomial over $F$, $N/F$ is a finite Galois extension. Identifying $\sigma$ with $\sigma \mid_{N}$ wish to show that $\operatorname{Gal}(N/F)$ is the cyclic group generated by $\sigma$. Clearly we've $\langle \sigma \rangle \leq \operatorname{Gal}(N/F)$. Then by Artin's theorem $\operatorname{Gal}(N/N^{\sigma})=\langle \sigma\rangle$. Since $N^{\sigma}=L^{\sigma}\cap N=F\cap N=F$, we've $\operatorname{Gal}(N/F)=\langle \sigma \rangle$. Thus $E/F$ is Galois and in fact $N=E$, by definition of $N$. Therefore $\operatorname{Gal}(E/F)$ is the cyclic group generated by $\sigma$.