Let $p: P \to B$ be a principal $G$-bundle, and $\pi : E \to P$ a vector bundle with action of $G$ on $E$ such that $G$ acts by vector bundle isomorphisms and $\pi$ is equivariant. Is it always the case that we get a vector bundle $E/G \to B$? If not, what are the conditions which need to be satisfied?
I was thinking the main thing which needs to hold is that we need $G$-invariant trivializations for $E \to P$. To this end, consider the following: Choose coordinate charts $U \times V$ of $P$ and $V$ of $B$ such that the projection $p : P \to B$ has the form $(x,y) \mapsto y$. Let $S \subseteq P$ correspond to the transverse slice $\{0\} \times V = \{(0,y)\}$. Then $p(S)$ is open and $W = p^{-1}(p(S)) = \bigcup_{g \in G} g \cdot S$ is a $G$-invariant tubular neighbourhood of the $G$-orbit through $(0,0)$.
Now $U \times V$ can be chosen such that $E$ is trivializable over this open subset and $U \times V \cong A \times S$, for an open subset $A \subseteq G$, where the map $A \times S \to U \times V$ is given by $(g,s) \mapsto g \cdot s$. Then $E|_{(U \times V)} \cong (U \times V) \times \mathbb{C}^r \cong (A \times S) \times \mathbb{C}^r$. Using the last representation, we can define $G$-invariant local section $f_{i}(g,s) := g \cdot ((e,s),e_{i})$, for $g \in A$, where $e_{i} = (0,...,1,...,0)$. Then we extend this to $W$ by defining $t_{i}(g \cdot s) := g \cdot f_{i}(e,s)$ now for all $g \in G$. This is smooth because for $g_{0} \cdot S$ we have the neighbourhood $g_{0} \cdot (U \times V) \cong (g_{0} \cdot A) \times S$, then for $a \in A$, $t_{i}(g_{0}a \cdot s) = g_{0}a \cdot f_{i}(e,s) = g_{0} \cdot f_{i}(a,s)$.
If this works then we have $G$-invariant local frame $t_{i}$ over $G$-invariant open set $W$. On a double overlap where we have two such frames $t_{i}$ and $t'_{i}$, the transition function will also be $G$-invariant. Therefore we get well-defined transition functions on the base $B$.
