First of all, I recommend reading Section III.1 of Serre's "Galois Cohomology", particularly III.1.3 on pp. 123-124. In that terminology, the key issue is whether or not the automorphism functor, $\text{Aut}_{X/k}$ in Serre's notation, is representable by a finite type $k$-scheme $G$. Assuming that it is, then the base change of the morphism $(m,\text{pr}_2):G\times_k X\to X\times_k X$ to a field extension $L/k$ is the analogous morphism for the base change $L$-scheme, $X_L = X\times_{\text{Spec} k} \text{Spec} L$. For some (finite) Galois extension $L/k$, there is an isomorphism $X_L \cong \mathbb{P}^{n-1}_L$. For this $L$-scheme, the multiplication morphism $(m,\text{pr}_2)$ is a surjective morphism of finite type $L$-schemes. Since surjectivity can be checked after a faithfully flat base change, this implies that the original morphism $(m,\text{pr}_2)$ over $k$ is surjective. Finally, since this is a morphism of finite type $k$-scheme, every surjective morphism is automatically surjective on closed points: the fiber over every closed point is a nonempty closed subscheme of the domain, which is finite type, hence the closed subscheme contains a closed point of the domain.

Therefore, the main issue is representability of $\text{Aut}_{X/k}$. This is dealt with in many places. The key here is that the dualizing sheaf $\omega_{X/k}$ is anti-ample, i.e., $\omega_{X/k}^\vee$ is ample. The functor $\text{Aut}_{X/k}$ is very close to the functor $\text{Aut}_{X/k,\omega_{X/k}^\vee}$ "parameterizing" a $k$-automorphism $f$ together with an isomorphism between $\omega_{X/k}^\vee$ and $f^*\omega_{X/k}^\vee$. This functor is itself a closed subfunctor of the linear $k$-group of $k$-linear automorphisms of $H^0(X,\omega_{X/k}^\vee)$. There is a discussion of all of this in Section 2.1 of my article with de Jong, "Almost Proper GIT-Stacks and Discriminant Avoidance".

`$(m,\text{pr}_2):G\times_k X \to X\times_k X$`

is a surjective morphism of schemes? Or are you asking whether the induced map of sets of $k$-points is surjective, $G(k)\times X(k) \to X(k)\times X(k)$? Of course the answer to both questions is yes, but the second question is a bit silly since $X(k)$ is nonempty if and only if $X$ is isomorphic to`$\mathbb{P}^{n-1}_k$`

. – Jason Starr Nov 16 '12 at 12:13