TITLE: Are all homogeneous metric spaces bihomogeneous?
QUESTION [8 upvotes]: Let (X,d) be a metric space such that for all points p and q in X, there exists an isometry f such that f(p) = q.  Does it follow that for all points p and q in X, there exists an isometry f such that f(p) = q and f(q) = p?
This seems like an obvious enough question that I would be surprised if the answer isn't simply a reference, but I haven't found it mentioned anywhere.

REPLY [5 votes]: Here is a one-dimensional analogue of Richard's triangle
example, obtaining a counterexample in the set of
reals. Namely, replace every integer $n$ with two numbers at
fixed small distance $n\pm\epsilon$. One can suitably translate and reflect to realize
homogeneity, but there is no isometry swapping $\epsilon$ and $1+\epsilon$.