TITLE: Does homeomorphic to itself imply the same topology?
QUESTION [9 upvotes]: I know homeomorphim is an equivalence relation, which means a topological space will be homeomorphic to itself. However, does the converse hold? In other words, is it possible that a set with two different topologies can still be self-homeomorphic?
Thank you.

REPLY [2 votes]: To add to the other answers, it is actually possible for a finer topology to be homeomorphic to a coarser one. I.e. you can add open sets to a topology and yet preserve homeomorphism-type.

Example
An example discussed here is $\mathbb{Q}$ with the Euclidean topology $\tau_E$ and the "Sorgenfrey'' topology $\tau_S$ generated by the base $\{(a,b]\mid a,b\in\mathbb{Q}\}$ are actually homeomorphic. Clearly, $(a,b)$ is always in $\tau_S$ but $(a,b]$ is never in $\tau_E$. So $\tau_S$ is strictly finer than $\tau_E$ 
The spaces $(\mathbb{Q},\tau_E)$ and $(\mathbb{Q},\tau_S)$ are homeomorphic because they are both countable metrizable spaces without isolated points and all such spaces are homeomorphic to $(\mathbb{Q},\tau_E)$ by theorem in the link. 

Edit (simpler example):
Consider $\mathbb R$ with the topologies 
$$
\tau_1=\{A\mid A^c\text{ is a finite set of integers}\}\cup\{\varnothing\},
$$
$$
\tau_2=\{A\mid A^c\text{ is a finite set of even integers}\}\cup\{\varnothing\}.
$$
Clearly $\tau_2\subsetneq \tau_1$ and yet $x\mapsto 2x$ is a homeomorphism from $(\mathbb R,\tau_1)$ to $(\mathbb R,\tau_2)$.