For an algebraic variety X over an algebraically closed field, does there always exist a finite set of (closed) points on X such that the only automorphism of X fixing each of the points is the identity map? If Aut(X) is finite, the answer is obviously yes (so yes for varieties of logarithmic general type in characteristic zero by Iitaka, Algebraic Geometry, 11.12, p340). For abelian varieties, one can take the set of points of order 3 [added: not so, only for polarized abelian varieties]. For P^1 one can take 3 points. Beyond that, I have no idea.
The reason I ask is that, for such varieties, descent theory becomes very easy (see Chapter 16 of the notes on algebraic geometry on my website).
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I get that the answer is "no" for an abelian variety over the algebraic closure of Fp with complex multiplication by a ring with a unit of infinite order. Since you say you have already thought through the abelian variety case, I wonder whether I am missing something.
More generally, let X be any variety over the algebraic closure of Fp with an automorphism f of infinite order. A concrete example is to take X an abelian variety with CM by a number ring that contains units other than roots of unity. Any finite collection of closed points of X will lie in X(Fq) for some q=p^n. Since X(Fq) is finite, some power of f will act trivially on X(Fq). Thus, any finite set of closed points is fixed by some power of f.
As I understand the applications to descent theory, this is still uninteresting. For that purpose, we really only need to kill all automorphisms of finite order, right?