I am not sure that all automorphism groups of algebraic varieties have natrual algebraic group structure. 
But if the automorphism group of a variety has algebraic group structure, how do I know the automorphism group is an algebraic group.
 For example, the automorphism group of an elliptic curve $A$ is an extension of the group $G$ of automorphisms which preserve the structure of the elliptic curve, by the group $A(k)$ of translations in the points of $A$, i.e. the sequence of groups
$0\to A(k)\to \text{Aut}(A)\to G \to 0$ is exact, see [Springer online ref - automorphism group of algebraic variaties](http://eom.springer.de/a/a011740.htm#a011740_00c2). 
In this example, how do I know $\text{Aut}(A)$ is an algebraic group.



Why do automorphism groups of algebraic varieties have natural algebraic group structure?
It is not always true that the automorphism group of an algebraic variety has a natural algebraic group structure. For example, the automorphism group of $\mathbb{A}^2$ includes all the maps of the form $(x,y) \mapsto (x, y+f(x))$ where $f$ is any polynomial. I haven't thought through how to say this precisely in terms of functors, but this subgroup morally should be a connected infinite dimensional object, and is thus not a subobject of an algebraic group.


On the other hand, I believe that the automorphism group of a projective algebraic variety, $X$, can be given the structure of algebraic group in a fairly natural way. This is something I've thought about myself, but not written down a careful proof nor found a reference for: For any automorphism $f$ of $X$, consider the graph of $f$ as a subscheme of $X \times X$, and thus a point of the Hilbert scheme of $X\times X$. In this way, we get an embedding of point sets from $\mathrm{Aut}(X)$ into $\mathrm{Hilb}(X\times X)$. 


I believe that it should be easy to show that (1) $\mathrm{Aut}(X)$ is open in $\mathrm{Hilb}(X\times X)$, and thus acquires a natural scheme structure and (2) composition of automorphisms is a map of schemes.


