This question is about a result in the section 16.3 of the book Linear Algebraic Groups from Humphreys.
The follow can be deduced from a proposition in the section 16.3 of the book:
Corollary: Let $G$ be a diagonalizable algebraic group, $V$ a connected variety and $\varphi:V\times G\rightarrow G$ a morphism of varieties. Assume also the map $x\mapsto \varphi_x$, given by $\varphi_x(y)=\varphi(x,y)$, is a group homomorphism for each $x\in V$. Them the map $x\mapsto \varphi_x$ is constant.
The book says its implies a diagonalizable group is 'rigid', i.e., there are few automorphisms. My question is if the following application is correct.
Example: Let $G$ be a diagonalizable algebraic group and $\varphi:\mathbb P^1\times G\rightarrow G$ a morphism of varieties with $\varphi_x\in Aut(G)$ (automorphisms of algebraic groups) for each $x\in \mathbb P^1$. Them by the corollary $x\mapsto \varphi_x$ is constant and there is only one automorphism indeed. This shows not exist a 'one parameter family' of automorphisms of $G$.
Yes, your application is correct: you just specify the corollary to $V=\mathbb{P}^1$.