Suppose $X$ is a smooth projective variety over a field $k$, with ample canonical bundle. If $\operatorname{char}(k)=0$ or $\operatorname{char}(k)>\dim(X)$ and $X$ lifts to $W_2(k)$ (thanks Olivier Benoist!), it's not hard to see that $X$ has finite automorphism group.

**Proof.** Let $X\hookrightarrow \mathbb{P}\Gamma(X, \omega_X^{\otimes n})^\vee$ be a closed embedding. Then the induced map $\operatorname{Aut}(X)\hookrightarrow PGL(\Gamma(X, \omega_X^{\otimes n})^\vee)$ is a closed embedding, so $\operatorname{Aut}(X)$ is finite type. But the tangent space to $\operatorname{Aut}(X)$ at the identity is $\Gamma(X, T_X)$, which is trivial by Kodaira vanishing. So $\operatorname{Aut}(X)$ is a $0$-dimensional finite type group scheme, hence finite. $\blacksquare$

(An essentially identical argument avoiding the representability of $\operatorname{Aut}(X)$ may be found here.)

This proof breaks down for $\operatorname{char}(k)$ small: $\Gamma(X, T_X)$ need not be trivial. But I don't know of an example of a canonically polarized variety with infinite (thus necessarily positive-dimensional) automorphism group in small characteristic.

Is the automorphism group of a smooth variety with ample canonical bundle always finite, even if $\operatorname{char}(k)$ is small?