I have been a bit sloppy in the title, but let me be specific. I stepped again into the subtle difference between homotopy limit and limit in the homotopy category, in the following version.

Suppose you have two diagrams of spaces $X_{\bullet}, Y_{\bullet}$ and a homotopy equivalence $f_i : X_i \to Y_i $ such that for any morphism $i\to j$, the morphisms $Y(i\to j)\circ f_i$ and $f_j\circ X(i\to j): X_i\to Y_j$ are homotopic. Is it true that $\text{holim} X_{\bullet} \simeq \text{holim} Y_{\bullet} $ ?

My favorite diagram is $\Delta$, i.e. cosimplicial objects, but I suspect the result should not depend on this. Let me remark that if the $f_i$ would commute strictly the result would be true.

**Bonus question.** If the answer is "No, they can be different", are there a higher compatibility constraints one can impose so that it becomes true? For example, by specifying what the homotopy $f_{ij}$ between $Y(i \to j) f_i $ and $f_j X(i \to j)$ should respect, and then bla bla..

**Remark**. I suspect that a "not so higher" version should be true for the following reason. If I have a cosimplicial object, I take the chain complex pointwise, and I apply the Dold-Kan correspondence, I get a bicomplex $C_*(X_\bullet)$: the horizontal differential is the differential of chains, while the vertical differential is the alternated sum of the cosimplicial (co)faces.
I have tried to apply the definition of multicomplex homotopy equivalence between $C_*(X_{\bullet})$ and $C_*(Y_{\bullet})$, but there a few more maps to find and I don't want to mess with so many equations. In particular I should find an homotopy inverse to $f$. I hope that the "not so high" is low enough that one homotopy suffices, without higher constraints.