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](http://math.stanford.edu/~dlitt/briefnotes/notes/curve-automorphisms.pdf).)


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.



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Do varieties with ample canonical bundle have finite automorphism group in small characteristic?
A much more general result is proven in [Martin-Deschamps, Lewin-Menegaux : Applications rationnelles séparables dominantes sur une variété de type général](http://archive.numdam.org/ARCHIVE/BSMF/BSMF_1978__106_/BSMF_1978__106__279_0/BSMF_1978__106__279_0.pdf). Théorème 2 : if $X$ and $Y$ are smooth and proper, and if $Y$ is of general type, the set of dominant separable rational maps from $X$ to $Y$ is finite !
 What you want follows immediately: it is their Corollaire 4, and they attribute it to Matsumura.


