Let $G\hookrightarrow P \xrightarrow{\pi} M$ be a principal bundle, denote by $\cdot$ the right action of $G$ on $P$. Let $f:P\rightarrow P$ a bundle automorphism (i.e. $f$ is a diffeo, $f(p \cdot g) = f(p)\cdot g$, $\pi\circ f=\pi$).
Is it true that with respect to a local trivialisation the action of $f$ is given by the left multiplication on the fibres?
The reason why I think so is the following one.
Let $(U,\Phi)$ be a local trivialisation, $\Phi:\pi^{-1}(U) \rightarrow U\times G$, $\Phi(p)=(\pi(p),\phi(p))$. Since $f$ does not move base points, $\Phi\circ f\circ \Phi^{-1}(x,e) = (x,h(x)) $ for some $h(x)=\phi(f(\Phi^{-1}(x,g))\in G$. Then $\Phi\circ f\circ \Phi^{-1}(x,g) = \Phi( f(\Phi^{-1}(x,e)\cdot g)=(x,\phi(\Phi^{-1}(x,e)) g)=(x,h(x) g) $ and with respect to the local trivialisation the action of $f$ is given by left multiplication on the fibres by $h:U\rightarrow G$.
Is the above correct?
Yes, it is correct. What you are thinking has a name (especially in physics), it is called a gauge transformation. See for example here and here.
Your reasoning is by the way a very nice way to see it.