Let $f:X\rightarrow Y$ be a smooth submersion of manifolds. Let $\xi$ be a vector field on $X$. We say that $\xi$ is projectable if there a vector field $\eta$ on $Y$ such that $df\circ \xi=\eta \circ f$.
I want to show that if $\pi:P\rightarrow M$ a principal $G$ bundle, $\xi$ a vector field on $P$ is projectable if and only if $(R_g)_*\xi=\xi$.
Suppose that $\xi$ is projectable. Then $d\pi \circ \xi = \eta \circ \pi$. The right hand side $\eta \circ \pi$ only depends on the fiber of $P$, and we stay in the same fiber when we act by some $g$. But I don't get to see from this that $(R_g)_*\xi=\xi$.
This is not true, take the trivial bundle $M\times \mathbb{R}^n$ and any vector field $U$ of $M$, consider a vector $V$ of $\mathbb{R}^n$ which is not invariant by the translation for example $ V(x_1,...,x_n)=(x_1^2,..,x_n^2)$ and $W$ defined by on $M\times\mathbb{R}^n$ by $W(x,y)=(U(x),V(x))$, $W$ is projectable since $d\pi W=U\circ \pi$ but is not invariant by the right translation.