$\operatorname{Aut}(G)=G(k)$ for a smooth algebraic group over $k$

Question

Let $G$ be a smooth algebraic group over a field $k$. Then $Aut(G)=\mathbf{G}(k)$, where $\mathbf{G}$ is the trivial right $G$-torsor.

My attempt:

Given a $k$-rational point $\operatorname{Spec}(k) \xrightarrow{g} G$, then one obtains an automorphism of $G$ by composing the structure morphism $G \rightarrow \operatorname{Spec}(k)$ with $g$.

For the other direction I am getting confused, and haven't been able to come up with a decent approach.

Any help please?

Answer 1

That is not true. For example $\textrm{Aut}(SL_2) = PGL_2(k)$. In any linear algebraic group, $\textrm{Inn}(G) \cong (G/Z)(k)$, where $Z$ is the center, and the full automorphism group is an extension of that by it's outer automorphisms.

In view of the edit:

An automorphism or torsors $ \varphi: \mathbf{G} \to \mathbf{G}$ is $G$-equivariant, hence $\varphi(g)=\varphi(e\cdot g)=\varphi(e)\cdot g$ where "$\cdot$" denotes right multiplication. Following the identity element, what can you conclude?