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][1]. According to [this post][2], 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.

  [1]: http://www.numdam.org/item?id=SB_1960-1961__6__249_0
  [2]: https://mathoverflow.net/q/22553/38052