Let $G\hookrightarrow P\xrightarrow{\pi}M$ be a principal $G$-bundle with right action $\cdot $ and suppose we are also given a left action $\rho: U\times P\rightarrow P$ of some group $U$ on $P$. Everything is smooth. Note that since left and right actions commute, for $u\in U$ the map $\rho_u=\rho(u, ):P\rightarrow P$ maps fibres to fibres and induces a map $\bar \rho_u$ on $M$. I am interested in the case where $\bar\rho_u$ is not the identity on $M$.
If $V$ is a vector space, $\kappa$ a representation of $G$ on $V$ and $E=P\times_\kappa V$ is the associated vector bundle with projection $\pi_E$, does $\rho$ induce in any natural way an action $\hat \rho$ of $U$ on $E$ so that the map $\hat\rho_u:E\rightarrow E$ is a vector bundle map (that is $\hat\rho$ maps fibres to fibres and $\hat\rho:E|_{\pi_(p)}\rightarrow E|_{\pi_(\rho_u(p))} $ is linear) which covers $\bar\rho_u$?
It seems to me that a suitable map $\hat\rho_u$ would be of the form \begin{equation} \hat\rho_u([p,v])=[\rho_u(p),L_u(p)v], \end{equation} with $L_u(p):V\rightarrow V$a linear map. In order for $\hat\rho_u$ to be well defined we need $L_u(p\cdot h)=\kappa(h)^{-1}L_u(p)\kappa(h)$. The action on the fibres of $E$ is linear and $\hat\rho_u$ maps fibres to fibres and covers $\bar\rho_u$ as \begin{equation} \pi_E(\hat\rho_u([p,v]))=\pi_E([\rho_u(p),v])=\pi(\rho_u(p))=\bar\rho_u(\pi(p))=\bar\rho_u(\pi_E([p,v])). \end{equation}
In general it seems to me that the only "preferred" choice for $L_u$ is to take $L_u(p)=\mathrm{Id}_V$ (or a multiple thereof) for all $p$, so that \begin{equation} \hat\rho_u([p,v])=[\rho_u(p),v]. \end{equation} If $U$ is a subgroup of the centre of $G$ we can also take $L_u(p)=\kappa(u)$, still independent of $p$.
First, am I saying anything stupid? And second, are there other natural ways of defining an induced action of $U$ on $E$ with the required properties?
EDIT: I have found the following related, but not equivalent question Group actions and associated bundles asking about induced actions on sections of an associated bundle. Sadly it is without answers too...
