By Cantor's Isomorphism Theorem, we know that any two given countable dense subsets $D_{1}, D_{2} \subseteq (0,1)$ are order isomorphic. However, this order isomorphism may be quite irregular in the sense that the image of $(0,\varepsilon)\cap D_1$ may have a diameter smaller than $\varepsilon$ while, for some $\varepsilon'<\varepsilon$, the image of $(0,\varepsilon')\cap D_1$ may have a diameter bigger than $\varepsilon'$. That is, around $0$, it may be contracting at some scale but expansive at a smaller one, hence irregular at zero. So I wonder
Q: What's the infimum $\nu_0\geq0$ such that, for any two given countable dense subsets $D_1,D_2\subseteq(0,1)$ and any $\nu\geq\nu_0$, we can find an order isomorphism $f_\nu:D_1\to D_2$ such that $\forall t\in D_1:f_\nu(t)\leq (1+\nu)t$? Is it a minimum?
If no such $\nu\geq0$ exists, we consider $\nu_0=+\infty$ as the infimum of the empty set. I believe that $\nu_0$ should be $0$. But I'm not convinced if it should be a minimum or not, i.e. if we can always find and order isomorphism $f: D_1\to D_2$ such that $\forall t\in (0,1):f(t)\leq t$. It could certainly be extended as an order isomorphism $f^\sharp:(0,1)\to(0,1)$ [and thus a homeomorphism from $(0,1)$ to itself] with $t\mapsto\sup\{f(s):s\in D_1\cap(0,t]\}$ where $f^\sharp$ would also satisfy $\forall t\in(0,1):f^\sharp(t)\leq t$ but this doesn't seem to pose any thread (a priori). So I don't know if there could exist some strange counterexample.