Galois Groups of Finite Extensions of Fixed Fields

I am trying to prove the following proposition:

Let $L$ be an algebraically closed field, $g \in Aut(L)$ and $K=\{x \in L \; | \; g(x)=x\}$. Show that every finite extensions $E/K$ is a cyclic Galois extension.

So far my thoughts have led me to the following:

If $E=K$, then clearly $E/K$ is Galois and $Gal(E/K)=Aut(K/K)=i$, where $i$ is the identity automorphism, so it is trivially cyclic. Assume $[E:K]=n$ for some $n>1$. Since $E/K$ is a finite extension, it is algebraic. Note that if $\beta \in E\setminus F$ then we have the minimal polynomial $m_{\beta}(X)$ over $K[x]$. Since $K$ is fixed by $g$, we know that $g(\beta)$ is also a root of $m_{\beta}(X)$. We can also conclude that $g(\beta)\not \in K$ since $\beta \not \in K$.

Where to take any of these observations, I have yet to determine. I feel I must be missing something elementary, for I cannot see how to conclude this is a Galois extension from here.

Any tips would be greatly appreciated, especially over full solutions.

Hint: Assume that $x(\in L)$ is algebraic over $K$. The set $\{g^i(x)\mid i\in\mathbf{N}\}$ is then finite, as $x$ has only finitely many conjugates over $K$. It follows that a suitable polynomial of the form $$ \prod_{i=0}^{n-1}(T-g^i(x))\in K[T]. $$ Using this you can hopefully prove that if $N$ is a normal closure of $E/K$, then $Gal(N/K)=\langle g\rangle$. The rest is easy.