Another tact; consider the set of foldings of our convex shape. Since the fold axis partitions the perimeter into two parts, and by convexity both have finite length, so one of them is at least half the length of the total perimeter.
edit: this needs fixing/to be made more precise --- see comments below. end edit
I claim * (without proof) that the longer part determines the congruence of the image with a subset of $F$, up to a Euclidean symmetry of $F$; whence w.l.o.g., we may assume the folding map fixes the longer part. Then we clearly require that the shorter part be mapped into $F$ --- in fact, it must be mapped to the intersection of $F$ with the half-plane containing the longer part.
Since we are interested in all such foldings, we may consider in particular the foldings that exactly bisect the perimeter; but then by oposite containments, such a folding must be a congruence of the two parts, and hence the folding axis is in fact a symmetry axis of $F$. An intermediate value theorem argument shows that there is such a folding map with axis in any direction, so $F$ has at least a circular group of symmetries; that is, $F$ is a circle.