Let the map-of-France shape M be the outer boundary of a (limiting) flowsnake: M_0 = a 2D regular hexagon, including its boundary curve. M_(n+1) = a rosette of 7 copies of M_n, after the rosette has been shrunk to total volume = 1 (linear factor of 1/sqrt(7)) and a slight rotation ... ... such that lim M_n = M. Since at each stage the M_n tessellate R^2, so does M. Call this tessellation by T_0. Then each tile of T_0 (a copy of M) is a rosette of 7 smaller ones. Call the tessellation by *those* by T_1, lather, rinse, repeat. Since a tessellation is nothing more than the set whose members are its tiles: Finally let T denote the set of all tiles of any T_n: T = T_0 u T_1 u ... u T_n u ... Whew! Now consider the *symmetry group* of all similarity mappings taking *any tile in T* to *any tile (possibly the same one) in T*, in *any rotational position* (there are 6). (Flips are excluded, of course, since all of M's isometries with itself are orientation preserving.) This is a very interesting group that I know little about. * * * What is the 1-dimensional version of this? That looks like a pretty interesting group G as well. A member of this group G takes any interval (K/2^n, (K+1)/2^n) to any interval of the same form, i.e., (L/2^m, (L+1)/2^m): (K/2^n, (K+1)/2^n) —> (L/2^m, (L+1)/2^m) by either preserving or reversing orientation and expanding length by a constant factor, i..e., by the map x |—> ax + b determined by (y - (L+1)/2^m) / (x - L/2^m) = ((L+1)/2^m - (K+1)/2^n) / (L/2^m - K/2^n). —Dan