Suppose $E\to X$ is a fibre bundle where say $X$ is a CW complex, with structure group $G$, i.e. it is classified up to bundle isomorphism by a homotopy class of maps $c\colon X\to BG$. It's not uncommon that a bundle can be non-trivial (i.e. $c$ is essential) but still the induced map $c^*\colon H^*(BG;R) \to H^*(X;R)$ is $0$ for any coefficients $R$, so that $G$ characteristic classes with values in singular cohomology do not detect the non-triviality of $E$.
My question is wether we can reduce to some other structure group $H$ and detect the bundle with $H$ characteristic classes. More specifically:
Given an essential map $c\colon X \to BG$ can we find a group homomorphism $\varphi\colon H\to G$ and a lift $\tilde{c}$ such that $$ 0\neq \tilde{c}^* \colon H^*(BH;R) \to H^*(X; R) $$ for some R?
(This question is related to one I asked here and an answer I gave here, except instead of creating an exotic cohomology theory I want to try reducing the structure group.)
The motivating example is $TS^{2n}$: all of the classes in $H^*(BO(2n);R)$ are stable and $TS^{2n}$ is stably trivial (in the sense of bundle-stabilization) so its $O(2n)$ characteristic classes all vanish. But, an orientation on $TS^{2n}$ induces a (homotopy class of) lift to $BSO(2n)$ which is detected by the Euler class. A case that I don't know how to settle is $TS^{2n+1}$, since I don't know about any characteristic classes with values in $H^*(-;R)$ that can detect when it is non-zero.
One snag when trying to detect bundles using characteristic classes is that if a classifying map is stably null-homotopic (in the sense of stable homotopy theory) then the induced map on any cohomology theory will be $0$. Therefore we have the following subproblem to deal with this case:
Given a map $c\colon X \to BG$ which is essential but stably null-homotopic, is there a $\varphi\colon H\to G$ and a lift of $\tilde{c}$ to $BH$ which is stably essential?
But even assuming the problem is reduced to the case that $c$ is stably essential I'm not sure how to proceed. The case of $SO(2n)$ characteristic classes detecting $TS^{2n}$ seems sort of like an accident since $SO(n)$ is nothing more than the identity component of $O(n)$. Likewise $Spin(n)$ is the universal cover of $O(n)$ (as long as $n\geq 3$) but eventually the connected covers stop being groups, and it's not clear where else to look. It's very possibly that the answer to my question is "no".
(Almost) yes, but this is kind of trivial and tautological: up to homotopy, every (connected) space is a classifying space, so every map classifies a bundle. To be precise, given any connected CW-complex $X$, there is a homotopy equivalence $X\to B\Omega X$ identifying $X$ with the classifying space of its loopspace $\Omega X$ (which is a group up to homotopy, but can be modeled up to homotopy equivalence with an actual topological group if that matters to you).
So, given a map $c:X\to BG$, we can always lift it to the identity map $X\to X$, which up to homotopy can be identified with our homotopy equivalence $X\to B\Omega X$. In your notation, we can consider this as a lift of $c$ with respect to the group homomorphism $\varphi=\Omega c:\Omega X\to \Omega BG\simeq G$, whose induced map $B\varphi=B\Omega c$ on classifying spaces is just $c$ itself. In particular, this lift is nontrivial on cohomology iff the cohomology of $X$ is itself nontrivial (which is of course a necessary condition for a lift which is nontrivial on cohomology to exist).
If you restrict your attention to bundles with structure groups that are (say) Lie groups, you probably get a more interesting question which I have no idea how to answer.