# Are all homogeneous metric spaces bihomogeneous?

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.

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The vertices of a snub cube form a metric space with 24 points that is homogeneous but not bihomogeneous: the edges of the squares have a "direction" associated with them.

Added later: here is an example with just 6 points: take an equilateral triangle with sides of length 1, and take the 6 points on the edges that are distance 1/4 from a vertex.

Added later: There are no examples with less than 6 points; for example, for 5 points there are 10 edges so there are at most 2 possible lengths with 5 edges of each length, which gives essentially only 1 configuration and this is bihomogeneous. Less than 5 points is easy to do case by case.

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So, how do we get the "direction" from the metric? – Ricky Demer Sep 28 '10 at 2:50
+1 for your excellent triangle example. Is it possible to beat 6 points? – Joel David Hamkins Sep 28 '10 at 3:15

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 $1+\epsilon$.

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You can make it slightly more symmetric by choosing pairs of the form $n\pm\epsilon.$ – Victor Protsak Sep 28 '10 at 3:42
OK, I changed it... – Joel David Hamkins Sep 28 '10 at 3:56
Both in Richard's example and your example, the isometry group, which is a dihedral group, acts simply transitively on the set of points. – Victor Protsak Sep 28 '10 at 4:16