Suppose $X\neq \emptyset$ is a set. Let $\tau_1, \tau_2$ be Hausdorff topologies on $X$ with the property that the partially ordered sets $(\tau_1,\subseteq)$ and $(\tau_2,\subseteq)$ are order-isomorphic.
Does this imply that $(X,\tau_1)\cong (X,\tau_2)$ as topological spaces?
If Hausdorffness is dropped, then there are easy counterexamples, for instance $X = \omega$ and $\tau_1 = \{\emptyset,\{0\},\omega\}, \tau_2 = \{\emptyset, \omega\setminus\{0\},\omega\}$.
If $h:(\tau_1, \subseteq) \to (\tau_2, \subseteq)$ is an order isomorphism.
Define $M(\tau_1) = \{O \in \tau_1: O \neq X \land \forall O' \in \tau_1: O \subseteq O' \implies O' = X\}$, the "submaximal" elements.
Because the definition is purely order theoretical, $h[M(\tau_1)] = M(\tau_2)$.
Then all $O \in M(\tau_1)$ are of the form $X\setminus \{p\}$ for some $p \in X$ when $X$ is a $T_1$ space: if $|X \setminus O| \ge 2$ pick $x$ in this complement, and $O \subseteq X\setminus \{x\} \subseteq X$, where the middle set is open by $T_1$-ness, this shows $O \notin M(\tau_1)$. So $\forall O \in M(\tau_1): |X \setminus O| =1$. Also, all sets of the form $X\setminus \{p\}$ are in $M(\tau_1)$.
So the above implies that if both topologies are $T_1$, we can define $f: X \to X$ by $\{f(x)\} = X\setminus h(M\setminus \{x\})$.
One can now check that $f$ is the required homeomorphism, using that $h$ is an order isomorphism.
There is more than an implication of a homeomorphism but actually an induced homeomorphism. It is so for all $T_1$-spaces.
Let $\ (X\ \tau)\ $ be a $T_1$-space. Then points of $\ X\ $ are characterized by the topology (treated as a Birkhoff lattice) as maximal elements of $\ \tau\ $ which are strictly contained in $\ X$.
