A metric space $M$ is *metrically homogeneous* if for every pair of points $x, y \in M$ there is an isometry $f$ of $M$ onto $M$ such that $f(x)=y$. What is known about metrically homogeneous spaces? Any references? In particular, is there an explicit description of all metrically homogeneous subspaces (under the Euclidean metric) of the plane $\mathbb{R}^2$?


**Remark.** The same question in one dimension has an easy answer: A subset $M$ of $\mathbb{R}^1$ is metrically homogeneous if and only if $M$ is the union of two cosets $G+a\ $ and $G+b\ $ of an additive subgroup $G$ of $\mathbb{R}^1$.



Metrically homogeneous subsets of the plane
Here is an description of homogeneous subsets of ${\mathbb R}^2$ using group theory, similar to the one you have in dimension 1. Let $M\subset {\mathbb R}^2$ be homogeneous and let $G< I=Isom({\mathbb R}^2)$ be its group of isometries. Then $G$ fits into short exact sequence
$$
1\to T\to G\to S \to 1
$$
where $T$ is a group acting by translations on the plane and $S$ is a subgroup of $O(2)$. Unlike in 1-dimensional case, this sequence need not split, but you can still find a subset $C\subset G$ of coset representatives of $S$ so that $G=TC$ (I am using action on the left). Then $M=G\cdot m$ for some $m\in M$ and, hence, 
$$
M= T\cdot (Cm),
$$ 
i.e., $M$ is a disjoint union of $T$-orbits, where $T$ is a subgroup of ${\mathbb R}^2$. This is an analogue of the description that you liked in the 1-dimensional case, except now $S$ is typically infinite. Thus, you have a (typically) infinite disjoint union, indexed by the set $S/G\_m$, where $G\_m$ is the projection of the stabilizer of $m$ in $G$ to the group $S$. I think, this is the best one can do without introducing further restrictions on the set $M$. Even in the case when $S$ is a finite dihedral group, there will be infinitely many possibilities for choosing a subgroup $T\subset {\mathbb R}^2$ normalized by $S$. 


In order to get something more geometrically appealing, let's assume that $M$ is a closed subset of ${\mathbb R}^2$. This immediately implies that the group $G$ is a closed subgroup of the Lie group $I=Isom({\mathbb R}^2)$. Basic Lie theory tells you that $G$ is a Lie subgroup of $I$. Now, the problem essentially reduces to classification of Lie subgroups of $I$. Here it is:


1. $dim(T)=2$, then $T$ acts transitively on ${\mathbb R}^2$ and, hence, $M={\mathbb R}^2$.
2. $dim(T)=1$. Then:


a. Either $T$ is either isomorphic to ${\mathbb R}$, acting via translations along a line $L\subset {\mathbb R}^2$, or 


b. $T\cong {\mathbb R}\times {\mathbb Z}$, where the first factor acts by translations along a line $L$ and the second acts by translations in the orthogonal direction. 


In either case, clearly, $M$ is a product ${\mathbb R}\times M\_1$, where $M\_1$ is a discrete homogeneous subset of the real line (there are four possibilities for $M\_1$ which you already know: single point, two points, orbit of the discrete group of translations or union of two such orbits). 


`3.` $dim(T)=0$. Thus, $G$ is a discrete group of Euclidean isometries. 


There are again 3 subcases here, depending on the rank of the free abelian group $T$ ($0, 1$, or $2$). If $T=0$ then $G< O(2)$ and, hence, $G$ is either 1-dimensional (in which case $M$ is a round circle) or finite cyclic or dihedral group; in the finite case everything reduces to vertex sets of regular or semiregular planar convex polygons. If $rank =1$ then $S$ is a subgroup of $Z\_2\times Z\_2$ and $M$ is either contained in one line or in two parallel lines. 


The most interesting case is when $T$ has rank 2 and, hence, $G$ is a Euclidean crystallographic group (Russians call them "Fedorov groups"). If you so desire, you can now go through the well-known list of such groups and identify $G$-orbits in ${\mathbb R}^2$. Some will give you esthetically pleasing vertex sets of regular and semiregular "floor-tiling" patterns on ${\mathbb R}^2$. If I were Joseph O'Rourke, I would add some nice pictures here, but I am not. 


