Metrically homogeneous subsets of the plane 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 $\mathbb{R}^1$.
 A: 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:


*

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

*$dim(T)=1$. 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. 
A: There's a complete classification that has been known since 1967: look past the Springer paywall here in the paper
Branko Grünbaum, L. M. Kelly. Metrically homogeneous sets, Israel Journal of Mathematics 
April 1968, Volume 6, Issue 2, pp 183-197.
If I recall correctly, there's a flaw in one of the lemmas as written, but a correction was published. I'll edit this answer once I find that correction.

Thanks to Francois Ziegler, we now have a link to the correction (see comment below). As pointed out in the other comments, the classification holds only for compact subsets of the plane. 

Also, as pointed out in the comments, the definition of metric homogeneity used in this paper differs from the OP's as follows: in their definition, a metric space $(M,d)$ is homogeneous if for each pair $x,y$ of points, the set of distances $d(x,z)$ equals the set of distances $d(y,z)$ when $z$ ranges over points in $M$. This is weaker than the requirement that there be an isometry taking $x$ to $y$, so the objects of interest to the OP are certainly a subset of the classification and I hope that the paper is at least a starting point: the literature seems rather thin aside from this.

On the other hand, let's take the paper's definition of metric homogeneity. Fix $x,y$ in a space $(M,d)$ satisfying this definition and note that there exists a set-valued map $F:M \to 2^M$ defined as follows: for each $z \in M$, define $F(z) \subset M$ by
$$F(z) = \{w \in M \mid d(w,y) = d(x,z)\}$$
I wonder what conditions one must impose on $(M,d)$ -- aside from the requirement that $M$ is a subspace of the Euclidean plane -- in order to guarantee that there exists an isometry $f:M \to M$ which is a selection of $F$, i.e., $f(z) \in F(z)$ for all $z \in M$. All the selection theorems which I can think of crucially require convexity of $F(z)$, but maybe there are experts who know other ways. 
