Can we detect non-trivial bundles using characteristic classes for a custom structure group?

Question

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".

Answer 1

(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.