Consider two points in $C$ with the largest distance. Clearly there must be a third that forms an equilateral triangle with these two. I claim that these three points form the convex hull of $C$ and there are no more points on the convex hull. Suppose not; then there exists a point $P$ that is outside of or on this triangle. Clearly $P$ must be in the reuleaux triangle determined by these three points or else it contradicts the maximality assumed at the start. Furthermore, let the vertex of this triangle furthest from $P$ be $A$. Then in order for $A, P$ to have a third point in $C$ that makes an equilateral triangle, this third point $P'$ must be $P$ rotated $60^{\circ}$ clockwise or counterclockwise w.r.t $A$. But clearly then $P'$ does not lie in the previous reuleaux triangle, which contradicts maximality of the distance of the two points first chosen. Now we know that any additional points must be strictly inside this maximal equilateral triangle. However, again if we have another point $P$ inside, then $A,P$ must have another point $P'$ in $C$ such that $P'$ is the image of $P$ under a $60^{\circ}$ rotation w.r.t $A$, but this means $P'$ does not lie in the equilateral triangle, contradiction. Thus, the only configuration possible is three points that form an equilateral triangle.

BUMP! The above solution is not complete.