Does the associated bundle functor have left or right adjoints? Let $\mathsf{Prin}_G$ be the category of (right) $G$-principal bundles, with a morphism from the bundle $p: P \to M$ to the bundle $p': P' \to M'$ being a pair of arrows $\chi: P \to P'$ and $\bar{\chi}: M \to M'$ such that $\chi$ is $G$-equivariant and the obvious diagram commutes. We only need to specify the morphism $\chi: P \to P'$ since the corresponding morphism down on the base is uniquely determined by $\chi$.
Let $\mathsf{Space}_G$ be the category of (left) $G$-spaces, with a morphism from $S$ to $S'$ being a $G$-equivariant arrow $f: S \to S'$.
Let $\mathsf{Bund}$ be the category of fibre bundles, with morphisms from $E \to M$, $E' \to M'$ being a pair of arrows such that we get a commutative square.
Then associated bundle assignment $$\mathsf{Prin}_G \times \mathsf{Space}_G \to \mathsf{Bund}$$
sending a pair of morphisms $\chi: P \to P'$ and $f: S \to S'$ to the morphism
$$\chi \times_G f: P \times_G S \to P' \times_G S'$$
given by
$$(\chi \times_G f) [u, s] = [\chi(u), f(s)]$$
is a bifunctor.
Now suppose we fix a principal $G$-bundle $P \to M$. Then this gives us an associated bundle functor $$P[-] := P \times_G (-): \mathsf{Space}_G \to \mathsf{Bund}(M),$$ where $\mathsf{Bund}(M)$ is the category of fibre bundles over $M$, with morphisms covering the identity on $M$.
Does this functor have any left or right adjoints? Do we need to restrict the target category to get an adjunction? What if instead we fix the other argument (the $G$-space) and let the principal bundle vary?
To provide a bit of motivation, I have read that for any fixed principal bundle the associated bundle functor from representations to vector bundles is exact, which leads me to guess that it might have both a left and a right adjoint. 
 A: I have at least managed to construct a left adjoint to the associated bundle functor 
$$P [-]: \mathsf{Space}_G \to \mathsf{Bund}(M),$$
Given by the "fibre product" functor 
$$ P \times_M (-): \mathsf{Bund}(M) \to \mathsf{Space}_G,$$
where the $G$ action is the obvious one $(u_x, v_x).g = (u_x.g, v_x)$. The counit is given by the $G$-equivariant maps
$$\varepsilon_S: P \times_M (P[S]) \to S$$
induced by the unique isomorphism $P \times_M (P[S]) \to P \times S$ (since both are pullbacks in the same pullback square) composed with the projection $P \times S \to S$. Explicitly, $\varepsilon_S(u_x, v_x) = q_{u_x}^{-1}(v_x)$, where $q_{u_x}: \{u_x\} \times S \to P[S]_x$ is the isomorphism given by restricting the quotient map $q: P \times S \to P[S]$. In the special case $S=G$, we find that the counit component $\varepsilon_G: P \times_M P[S] = P \times_M P \to G$ is just the usual division map for a principal bundle.
The unit map, however, seems a bit more elusive to describe, and I have not yet found a right adjoint to the associated bundle functor, if it exists.
As a nice application of this adjunction, we obtain the following classical result. Since the global sections functor $\Gamma: \mathsf{Bund}(M) \to \mathsf{Sets}$ is representable by the trivial bundle $\text{id}_M: M \to M$, we have a natural bijection 
$$\text{Hom}_{\mathsf{Space}_G}(P, S) \cong \Gamma(M, P[S])$$
which identifies the sections of the associated bundle $P[S] \to M$ with $G$-equivariant maps $P \to S$.
