Group action under homotopy equivalence. Suppose $\bar{X}$ is homotopy equivalent to $\bar{Y}$. Suppose the map $f_1:\bar{X}\rightarrow \bar{Y}$ and $f_2:\bar{Y}\rightarrow \bar{X}$ gives the homotopy equivalence. Suppose $G$ is a group acting discontinuously on $\bar{Y}$ and the quotient space is $Y$. Now we can define the action of $G$ on $\bar{X}$ as follows. Let $x\in \bar{X}.$ Then for $g\in G$ define $g(x)=f_2gf_1(x).$ Suppose under this action of $G$ on $\bar{X}$ the quotient space is $X$. 
Is this action will be discontinuous?
If so: 
Is $X$ is homotopy equivalent to Y.
If not 
Is there any condition (like properly discontinuous action) of the action will ensure the above facts to be true? 

This is not a group action.  In particular, $f_2f_1(x)$ need not equal $x$, so the identity element of $G$ doesn't act trivially.  Indeed, the only obvious case where you get an action is if $f_1$ and $f_2$ are an inverse pair of homeomorphisms.
Instead, the homotopy equivlance induces a group action on $\overline{X}$ up to homotopy.  That is, each element of $G$ corresponds to a homotopy class of maps $\overline{X}\to\overline{X}$.  Unfortunately, this isn't good enough to let you construct a quotient $X$.
