Geometric realization of a "simplicial space up to homotopy," part two

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This question is a follow-up, and my initial motivation for asking Is there a sensible way to form the geometric realization of a "simplicial space up to homotopy"? Given that the questions are essentially different, it seemed sensible to divide it into a separate post. In the hypothetical application I'm interested in, I don't actually want the operation nor the action on $X$ from the previous question to be totally defined. As a concession, I'll demand associativity of $G$ (because I don't think the simplicial set up to homotopy will have well-defined objects anymore otherwise: which elements are composable could depend on the bracketing).

Let $G$ be a topological category, $m$ its composition, $\pi\colon X \to G^{(0)}$ a space fibered over the objects, $G^{(1)} {}_s \!\times_\pi X$ the fiber product consisting of those pairs $(g,x)$ such that the source of $g$ is $\pi(x)$, and $$\mu\colon G^{(1)} {}_s \!\times_\pi X \to X$$ a partial "action" such that $\pi(\mu(g,x)) = t(g)$.

If we write $G^{(2)} \subset G \times G$ for the space of pairs of composable elements, then there are two natural maps $$G^{(2)} {}_s \!\times_\pi X \to X,$$ namely $ \mu \circ (\mbox{id} \times \mu)$ and $ \mu \circ (m \times \mbox{id})$. I want to assume these two maps are homotopic but not equal.

If the maps were equal, then $G^{(1)} {}_s \!\times_\pi X$ would be a well-defined topological category traditionally denoted $G \ltimes X$, with $s(g,x) = x$ and $t(g,x) = t(g)$. One could then form the nerve $N(G \ltimes X)$ (for example, we would have $N(G \ltimes X)_2 = G^{(2)} {}_s \!\times_\pi X$), and the geometric realization would be $B(G \ltimes X)$.

Is there a way to modify geometric realization to get a good analogue of $B(G \ltimes X)$ in this generality?

In particular, if $F \to X \overset\pi\to G^{(0)}$ were a fiber bundle, I would like this analogue to be, up to homotopy, an $F$-bundle over $BG$. I am happy to assume $G$ is a groupoid.