Let $A\subset \mathbb R^2$ be a set that has an axis of symmetry in every direction, that is, for any $n\in S^1$, there exists a line $D$ orthogonal to $n$, such that $A$ is invariant under the (affine) reflection of axis $D$.
It is easy to show that if all of these axes intersect at some point, say $O$, then $A$ is a reunion of concentric circles of center $O$. That is because the reflections generate the orthogonal group. Such a set is said to be radial.
If $A$ is assumed to be bounded, or to be Lebesgue-measurable and of positive and finite measure, then it must be radial. To see this, consider two orthogonal axes of symmetry. They intersect at some point $O$, and $A$ is invariant under the point reflection across $O$. Now consider two different orthogonal axes, that intersect at $C$; $A$ is also invariant under the point reflection across $C$. Thus, $A$ is invariant under the translation by $2\,OC$. This is only possible, of course, if $C=O$, which means all axes of symmetry intersect at $O$, and $A$ is radial.
I feel like the following statement is true, but have been unsuccesful in proving so.
If $A\subset\mathbb R^2$ has an axis of symmetry in every direction, then it is radial.
Can it be proved, or conversely, is there a counterexample?