Warning: I suspect that the answer to my following question is negative in this generality, however it seemed the best way to express what kind of results I am looking for (see my last point).
Suppose we have two homology theories on the same category, i.e. two covariant functors $$h_*, h_*' \colon \mathcal{L} \to \mathsf{Ab}$$ that factor through the category of chain complexes $\mathsf{Comp}$,i.e. $$h_*=H_*(C_{\bullet}) \text{ and } h_*'=H_*(C_{\bullet}'),$$ where $C_{\bullet}, C_{\bullet}' \colon \mathcal{L} \to \mathsf{Comp}$ are covariant functors. Suppose that $h_*$ and $h_*'$ are naturally isomorphic via a natural transformation $$\Phi_* \colon h_* \to h_*'.$$
My question is the following:
Can one recover information/any relations between the chain complex functors $C_{\bullet}, C_{\bullet}'$? Or more concretely:
Does there exist a natural isomorphism $$\Psi_{\bullet} \colon C_{\bullet} \to C_{\bullet}'$$ that coincides with $\Phi_*$ in homology?
(Edit: This does not hold in general, see comments below)
New question: Does there exist a natural chain map $\Psi_{\bullet}\colon C_{\bullet} \to C_{\bullet}'$ that agrees with $\Phi_*$ in homology?
Ordinary homology theories in the sense of Eilenberg-Steenrod together with the uniqueness result of such theories seem to be a good starting point, but I was not able to find anything useful (probably due to my lack of knowledge in (axiomatic) homology theory…).
Best Answer
The "natural" bit here is a red herring, and seeing that it's not relevant makes the question a lot easier.
Indeed at this level of generality ($\mathcal{L}$ an arbitrary category) we can let functors "be anything" and "natural" sort of loses its point. Indeed, it suffices to take $\mathcal{L}$ to be the discrete category on one object.
At that point, a functor $\mathcal{L}\to \mathsf{Comp}$ is just the choice of a complex, and $\mathcal{L}\to \mathbf{Ab}$ just the choice of an abelian group.
A natural isomorphism (or more generally transformation) is then just an isomorphism (or more generally morphism). Hence at this level of generality, a special case of your question becomes :
The answer to that is clearly no. For instance consider a complex $\mathsf{C}$ with only $\mathbb{Z/2Z}$ in position $0$ and $0$'s elsewhere, and $\mathsf{C'}$with $\mathbb{Z}$ in position $-1$ and $0$, the map $\mathbb{Z}\to \mathbb{Z}$ being multiplication by $2$ . Then the homologies of both complexes are isomorphic (they're $\mathbb{Z/2Z}$ in position $0$, and $0$ elsewhere); but there is no nontrivial morphism $\mathsf{C\to C'}$, hence no morphism that could induce the isomorphism in homology.
In fact this isn't category-dependent (you might argue that I chose a dummy category): for any category $\mathcal{L}$, you can "model" this situation by picking constant functors, and the result will be the same; therefore if you want to add some conditions to change the answer to "yes", then the conditions will not only be on $\mathcal{L}$ but also on the functors you have, e.g. respect products or whatever. But that would be another question, and right now I don't have examples of natural (non trivial) constraints one could add on the functors to have a positive answer.
The moral here is that the functor $H_* : \mathsf{Comp}\to \mathbf{Ab}$ is very very far from being full.