If a variety $X$ has finite automorphism group, is the same true for its $n$-fold self-products? Let $X$ be an algebraic variety over $\mathbb{C}$.
Let $n\geq 1$ be an integer and let $X^n$ be the $n$-fold self product of $X$.


Q. Is there an integer $n\geq 1$ and an algebraic variety $X$ over $\mathbb{C}$ such that $\mathrm{Aut}(X)$ is finite and  $\mathrm{Aut} (X^n)$ is infinite?


 A: The following result of Biswas and Gómez provides a partial positive answer (for the symmetric product) in the case where $X$ is a curve.

Theorem. Let $X$ be an irreducible smooth projective curve of genus $g > 2$ defined over an algebraically closed field of characteristic different from two, and take an integer $d > 2g-2$. Then the natural group homomorphism $$\mathrm{Aut}(X) \longrightarrow \mathrm{Aut}(\mathrm{Sym}^d X)$$
  is an isomorphism.

See 
I. Biswas, T. L. Gómez: 
Automorphisms of a symmetric product of a curve (with an appendix by Najmuddin Fakhruddin).
Doc. Math. 22 (2017), 1181–1192. MR3690270. 
A: This is partially answered in the answer to this question: - the question itself is different. The answer applies when you have a product of two curves, but as the answer-er (Roberto Pignatelli) points out, the key result (lemma 3.8 of Catanese's paper referenced) does not appear to rely on the variety being a curve).
Catanese, Fabrizio, Fibred surfaces, varieties isogenous to a product and related moduli spaces, Am. J. Math. 122, No. 1, 1-44 (2000). ZBL0983.14013.
A: For $X$ projective, we have that $\mathrm{Aut}(X)$ finite implies $\mathrm{Aut}(X^n)$ finite.
In this case, $\mathrm{Aut}(X)$ is representable by a group scheme of finite type (*Edit. locally of finite type)  over $\mathbf{C}$ and its Lie algebra is identified with global tangent fields $H^0(X, T_X)$, after Grothendieck. According to this post, the group scheme $\mathrm{Aut}(X)$ is reduced, so the condition that $\mathrm{Aut}(X)$ being finite is equivalent to that $H^0(X, T_X) = 0$. It then suffices to observe that $H^0\left(X^n, T_{X^n}\right) \cong H^0(X, T_X)^{\otimes n}$.
Edit Yes, a priori $\mathrm{Aut}(X^n)$ may be discrete and infinite. Sorry for that.
