Is the following true?
Let $X$ be a topological space and $\rho:X\to Y$ a quotient map onto a topological space $Y$. Suppose further that there exist a continuous function $f$ mapping $X$ onto itself with the property that whenever $C\subset X$ equals the preimage under $\rho$ of any subset $U\subset Y$, then $f(C)$ also equals a the preimage under $\rho$ of some subset $V\subset Y$ so that $U$ and $V$ are homeomorphic. Then there is a unique continuous map $g:Y\to Y$ so that the following diagram commutes
$$ \newcommand{\ra}[1]{\xrightarrow{\quad#1\quad}} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ll} X & \ra{f}& X\\ \da{\rho} & & \da{\rho} \\ Y & \ra{g} & Y \\ \end{array} $$
Let $y$ in $Y$ and choose some preimage $x$ of $y$ under $\rho$ in $X$. Now set $g(y)$ to be the image of $x$ under $$\rho \circ f : X \to X \to Y$$ this clearly makes the diagram commute, but we must show that it is well-definied.
Consider the sets $C = \rho^{-1}(\{y\})$ and let $V$ be so that $f(C) = \rho^{-1}(V)$ (which exists by hypothesis). Now $\{y\}$ and $V$ are homeomorphic, in particular bijective, so $V$ is a single point. Hence the map is well defined since any $x \in C$ maps into $V$.
If $U$ is an open subset of $Y$ then by continuity of $\rho$ and $f$, the preimage $\rho^{-1}(g^{-1}(U))$ is open. By definition of the quotient topology, $g^{-1}(U)$ is then open.
Uniqueness is clear since $\rho$ is surjective.