$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}$ Suppose we have $CW$-complex $X$. All the self-homotopy equivalences form a monoid, denote it by $G$.

Question: *is there any good way to construct another space $\widetilde X$, such that $\widetilde X$ is homotopy equivalent to $X$ and there exists a homomorphism from $G$ to the group $\widetilde G$ of all self-homeomorphisms of $\widetilde X$*?

A *good way* means that, (besides the functoriality) for any homotopy between elements of $G$ we should have an isotopy between corresponding elements of $\widetilde G$. Also, for every $f\in G$ the following diagram should be homotopy-commutative:

$$ \begin{array}{c} X & \ra{f} & X \\ \da{e} & & \da{e} \\ \widetilde X & \ra{\widetilde f} & \widetilde X\end{array} $$

Here $e$ is a fixed homotopy equivalence between $X$ and $\widetilde X$.

(I have one idea how to change all the homotopy equivalence to the homeomorphisms using mapping telescope, but I don't know what to do with homotopies and isotopies)

monoidinstead ofgroupoid, i forgot the right word. of course, compositions of homotopy equivalences are very important (and homotopies are important too, as it written below) $\endgroup$ – Andrey Ryabichev Aug 14 '16 at 11:50not homotopic, then the corresponding elements $\widetilde f,\widetilde g\in\widetilde G$ are not homotopic too. but this condition holds automatically, so the map $G\to\widetilde G$ should be an arbitrary homomorphism $\endgroup$ – Andrey Ryabichev Aug 14 '16 at 14:08