For context, I am working on my bachelor thesis on algebraic K-theory. I have "finished" the basics of algebraic topology, by which I mean the main content of Hatcher's book. For my project I need to be comfortable with twisted coefficients, which I am not yet.
In particular, suppose $X$ is a simple space, i.e the action of $\pi_1(X)$ on the higher homotopy groups is trivial. Let $M_\rho$ be a module twisted by $\rho: \pi_1(X)\to Aut(M)$, I want to understand $H_*(X; M_\rho)$. Intuitively, I feel like the trivial action of $\pi_1(X)$ on higher homotopy groups means that $H_*(X;M_\rho)\cong H_*(X; M)$. However the isomorphism I am almost certain doesn't hold at the chain complex level, and upon trying to prove that the isomorphism appears when we work on cycles modulo boundaries, the fact that I am not comfortable with what I am doing is making me hesitate that my proof works.
Any help in proving or disproving the statement is appreciated, thanks. (as an extra, I technically would be content with proving the result in the case that $X$ is an H-space, but curiosity and intution lead me to ask the question under the weaker hypothesis that $X$ is simple)
A simple space has the property that $\pi_1$ acts trivially on all homotopy groups, not just the higher ones, in particular $\pi_1$ is abelian. Otherwise there are very obvious counterexamples to your question because group cohomology regularly uses local coefficient systems to study $K(G,1)$ for nonabelian $G$. However, even with this added property of simple spaces, your question still has a negative answer. In particular, the homology $H_*(X; \mathbb{Z}[\pi_1(X)])\cong H_*(\widetilde X; \mathbb{Z})$ which will generally be nonisomorphic to what we would get with the trivial action on $\mathbb{Z}[\pi_1(G)]$. In particular, these are very different for the simple space $S^1$.