Let $G$ be an affine algebraic group, and let $X$ be a smooth variety with a $G$-action $\sigma:G\times X\rightarrow X$. Let $\pi: G\times X \rightarrow X$ denote the second projection. A $G$-linearized line bundle $L$ on $X$ is by definition a line bundle together with a bundle isomorphism $\Phi: \pi^*L\simeq \sigma^*L$ satisfying certain conditions.
However, it seems to me that a line bundle $L$ always has a canonical map $\pi^*L\rightarrow \sigma^*L$ of line bundles; namely, identifying $\pi^*L$ with $G\times L \rightarrow G\times X$, we can consider the map $G\times L \rightarrow L$ given by $(g,(x,v)) \mapsto (gx,v)$, and this map lies over the action map $\sigma:G\times X\rightarrow X$. Thus by universal property of fiber products, there should exist a map $\pi^*L\rightarrow \sigma^*L$.
My questions are as follows:
Is my reasoning correct, i.e. do we always have a natural map $\pi^*L\rightarrow \sigma^*L$ as above?
When we say that the line bundle $L$ has a linearization, do we mean that the natural map in 1. is an isomorphism?
Thanks in advance.