Why doesn't two topologies being homeomorphic imply that the induced map between the topologies is a homeomorphism? Suppose that $h : X \rightarrow Y$ is a homeomorphism. The homeomorphism $h$ induces a bijection $h_* : \tau_{\small X} \rightarrow \tau_{ \small Y}$ between the topologies of $X$ and $Y$. 
My question is: what do we mean when we say that the two topologies $\tau_{\small X}$ and $\tau_{\small Y}$ are homeomorphic? I understand that a homeomorphism $h$ between $X$ and $Y$ means that $X$ and $Y$ are homeomorphic as sets/spaces, but when we say that $\tau_{\small X}$ and $\tau_{\small Y}$ are homeomorphic, aren't we just saying that there is a homeomorphism between some subsets of $\mathcal{P}(X)$ and $\mathcal{P}(Y)$, respectively, which would mean that $h_*$ is not only a bijection, but also a homeomorphism itself? Or have I just got confused between the notion/definition of homeomorphic topologies?
Edit: Also, as I side question, I'm guessing that $h_*$ is the thing which maps open sets to open sets, not necessarily $h$. Is there an example of a homeomorphism $h$ which doesn't map open sets to open sets?
Edit 2: (from the comments) If we have a homeomorphism map  $h : X \rightarrow X$, and we give a topology $\tau_1$ to the domain, and a different topology $\tau_2$ to the codomain, is it possible that the domain and codomain won't be homeomorphic?
Many thanks for any answers.
 A: The answer to question $2$ is emphatically yes. Let $\tau_1$ be the indiscrete topology (only $X$ and $\emptyset$ open) and let $\tau_2$ be the discrete topology (every set open); it's easy to check that if $X$ has at least two elements, the identity map is not a homeomorphism (and indeed there is no homeomorphism).
Note, though, that the way you've phrased that question is a bit odd: the particular map $h$ doesn't come into play at all. I think what you meant was, "if $h$ is a map $X\rightarrow X$ and $\tau_1, \tau_2$ are topologies on $X$, is $h$ necessarily a homeomorphism from $(X, \tau_1)$ to $(X, \tau_2)$?" But note that this isn't quite what you asked.
A: My answers to your two final questions:


*

*No, there is no such an example. If $h$ is a homeomorphism and $A$ is an open subset of $X$, then $h^{-1}$ is continuous, and therefore $(h^{-1})^{-1}(A)$ is an open subset of $Y$. But $(h^{-1})^{-1}(A)=h(A)$.

*Yes, even if the map is continuous. For instance, take $X=\mathbb R$, with the discrete topology on the left and the usual one on the right.

