TITLE: Automorphism group of a variety
QUESTION [7 upvotes]: Suppose $X$ is a (quasi-projective) variety over a field $k$, and let $\mathbb{P}^n(k)$ be the ambient projective space. 

When can one decide that the automorphism group of $X$ is induced by a subgroup of the automorphism group $\mathbf{P G L}_{n + 1}(k)$ of $\mathbb{P}^n(k)$ ? 

When $X$ itself is a sub-projective space over $k$, this is of course well known, but when $X$ is e.g. an affine subspace over $k$, this is in general not true (due to the existence of non-linear automorphisms).

Is it true when $X$ is projective ?

REPLY [9 votes]: I guess this should be true if and only if $f : X \to X$ fixes a polarization, i.e. there is an ample $A$ such that $f^\ast A \sim A$.
If there is such an $A$, then take the embedding of $X$ into $\mathbb P^n$ by a very ample $mA$.  The automorphism of $X$ is then induced by the linear map $f^\ast : \mathbb PH^0(X,mA) \to \mathbb PH^0(X,mA)$.  Conversely, if a map comes from a linear map on projective space, then the polarization given by restricting the hyperplane is fixed.
Typically automorphisms won't fix any polarization, so don't come from $\mathbb P^n$.  If you actually want to check it for a specific map, the first thing to do work out the induced map on $H^2(X;\mathbb R)$.  If there is an invariant ample divisor, its first chern class will be invariant under the pullback.  So you may be able to show that this doesn't happen.  If there is a fixed class, you need to know (a) whether it is represented by an ample divisor, which may or may not be easy to check depending on what $X$ is and (b) whether the divisor is linearly equivalent (rather than just numerically equivalent) to its pullback, which requires a bit more scrutiny.
As Francesco points out, $K_X$ always pulls back to itself under an automorphism, so if it's either ample or antiample you have what you want.