Let $A_1,A_2,\ldots, A_n$ be distinct in the plane. For every $1\le i \le n$, let $S_i=\sum\limits_{j=1}^n d(A_i,A_j)$ be the sum of distances from $A_i$ to all the other points.
Assume that $S_i=S_j$ for all $1\le i<j\le n$. Is it true that the points $A_1,A_2,\ldots, A_n$ must be the vertices of a convex $n$-gon?
Suppose that the points are not in convex position, then there is a point (WLOG let it be $A_n$) that is a convex combination of other points:
$$A_n = \sum_{i=1}^{n-1} \lambda_i A_i\text,$$
where $\lambda_i\ge0$ and $\sum_i\lambda_i=1$. Now consider the function $f_i(X) = d(A_i,X)$. This function is convex: its value at a convex combination of points is no more than the same convex combination of the values at the points. It is strictly convex as long as the points are not on a line through $A_i$. Therefore, you have that
$$S_n = \sum_{i=1}^{n-1} d(A_i,A_n) < \sum_{i=1}^{n-1}\sum_{j=1}^{n-1} \lambda_j d(A_i,A_j) < \sum_{j=1}^{n-1} \lambda_j S_j\text.$$
This is impossible if $S_i = S_j$ for all $i,j$, so, by contradiction, the points must be in convex position.
