If $A$ and $B$ are countable dense subsets of compact metric space $(X, d)$ with infinitely many isometries, is there isometry such that $f(a) \in B$.

Question

Let $(X, d)$ be a separable and compact metric space with infinitely many isometires. Let $A, B$ be infinitely countable dense subsets of $X$. Is there an isometry $f : X \rightarrow X$ such that $f(a) \in B$, for some $a \in A$?

Isometry is a function $f : X \rightarrow X$ which satisfies $d(f(x), f(y)) = d(x, y), \forall x,y \in X.$

Answer 1

In general, no. Consider the cylinder $\mathbb S^1\times I\subseteq \mathbb R^3$, where $\mathbb S^1$ is the unit circle in $\mathbb R^2$, and $I=[0,1]$, with the metric inherited from the Euclidean metric on $\mathbb R^3$.

Then there are uncountably many isometries (every isometry of the circle extends to one, and one may also compose with reflections through the plane $z=\frac{1}{2}$), but if $A$ is the set of points at rational height and rational angle, and $B$ is the set of points with height and angle in $\sqrt{2}+\mathbb Q$, then $A$ and $B$ are countable and dense, and no isometry will take any point in $A$ to a point in $B$, since all isometries preserve height (or invert height, but in the latter case, rational heights are still sent to rational heights.)