Let $X$ be a smooth, rationally connected variety over an algebraically closed field of characteristic zero. Denote by $\mathrm{Aut}(X)$ the space of automorphisms of $X$ and for a given $\phi \in \mathrm{Aut}(X)$, denote by $\mathrm{Fix}(\phi)$ the fixed points of the automorphism $\phi$. I wanted to ask, if there is any known condition under which the intersection $F$ of all $\mathrm{Fix}(\phi)$ as $\phi$ ranges over all elements of $\mathrm{Aut}(X)$ is the emptyset?
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$\begingroup$ The identity component $G$ of the group scheme $\text{Aut}(X)$ is linear. If every $G$-orbit has stabilizer group equal to a proper subgroup, then there is no point fixed by all of $\text{Aut}(X)$. The dimension of the stabilizer group is the nullity of the fiber of the map $\mathfrak{g}\otimes \mathcal{O}_X \to T_X.$ There are plenty of other examples, e.g., an action of $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$ with nontrivial "elementary obstruction" has empty fixed locus, cf. mathoverflow.net/questions/284559/… $\endgroup$– Jason StarrOct 31, 2017 at 9:15
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