Let $K$ be an algebraically closed field, and let $\sigma$ be an automorphism on $K$. Set $k=K^\sigma$. Consider the rational function field $K(x)$ and extend $\sigma$ to $K(x)$ by $\sigma(x)=x$, hence $K(x)^\sigma=k(x)$.
Can every finite separable extension $L/K(x)$ with the property that $\sigma$ extends to an automorphism on $L$ be written as a compositum $L=Kl$ for some finite, separable extension $l/k(x)$?
In the special chase char$(K)=0$, this seems to follow from the fact that any covering over $K$ comes from a covering over $\overline{k}$ [Theorem 6.3.3 in Serre - Topics in Galois Theory].
