It seems to be a common fact that transition maps of principal bundles are given by multiplication by a group element (in this post,for example.)
The setup is this: Consider a $G$-bundle $E \overset{\pi}{\rightarrow} B$ and open sets $U,V$ with corresponding trivialisations
These induce trivialisations $\pi^{-1} (U \cap V) \overset{\varphi}{\rightarrow} (U \cap V) \times G$ and $\pi^{-1} (U \cap V) \overset{\psi}{\rightarrow} (U \cap V) \times G$, so we get a transition map
$$ \varphi \circ \psi^{-1} : (U \cap V) \times G \rightarrow (U \cap V) \times G$$
Why is this map necessarily given by (right) multiplication by an element of $G$?
It seems that the more general idea here is that any (equivariant) endomorphism of a $G$-torsor T (i.e. set on which $G$ acts freely and transitively) is given by multiplication by a group element (as seems to be stated in the first answer of this post)
So let $\rho: T \rightarrow T$ be such a map. Fix an element $x \in T$ and let $g$ be the unique element of $G$ such that $\rho(x) = x.g.$
For any $y \in T$, there exists an unique element $h \in G$ such that $y = x.h$. Then
$$\rho(y) = \rho(x.h) = \rho(x).h = (x.g).h \neq (x.h).g = y.g,$$
in general. So why is the transition map given by multiplication by an element of $G$?
(Continued from comments) The trivial bundle has a left and right action, and both are being used even if you want all torsors to be right torsors, the MAP between two trivializations is given by LEFT multiplication. More details say as a G torsors we consider UxG to be equipped with right G action. Let's for simplicity take U to be a point, so we have a map G->G that is right G-equivariant. So it is determined by where e goes to, say g. Then h=eh goes to gh, by G equivariance and all the as actions is on right. I.e the MAP (not the torsor actiob) is left multiplication by g as h goes to gh. If your group is abelian then this is also a right multiplication obviously. If you look at end of p.2 prop 2.1 of the pdf by Mitchell, you see f(b) is acting on the left.