Let $(X,\tau)$ be a topological space. For $x,y\in X$ we set $$\Delta(x,y) = \{(z,z):z\in X\}\cup \{(x,y),(y,x)\}.$$


Is there a connected Hausdorff space $(X,\tau)$ with more than $1$ point such that for all $x,y\in X$ we have $X\cong X/\Delta(x,y)$ with respect to the quotient topology?



Connected Hausdorff space that is stable under collapsing two points
Start with the space $X\_0=S^1$ (any connected manifold will do), and choose a basepoint in $X\_0$. Construct a sequence of pointed spaces recursively as follows: $$X\_{n+1}=\bigvee\_{x,y\in X\_n, x\ne y} X\_n/
\Delta(x,y).$$ Let $X=\bigvee\_{n=1}^\infty X\_n$.


Then $X$ is connected and Hausdorff. Moreover, $X$ is naturally identified with a wedge sum with the following properties: 1. For each wedge summand, there are infinitely many other summands homeomorphic to it, 2. for any two points in a wedge summand $A$, there is a wedge summand homeomorphic to the quotient of $A$ by the two points, 3. For any two points belonging to different wedge summands $A, B$, there is a wedge summand homeomorphic to the quotient of $A\vee B$ identifying these points. From here it is easy to conclude that $X$ has the property that you want.


