Let $S\subset \mathbb{R}^2$ be a bounded convex set. For which $S$ can we take $S'$ congruent to $S$ (i.e., the image of $S$ under an isometry of the Euclidean plane) such that $S\cap S'$ is similar to but not equal to $S$?
For instance, any triangle or quadrilateral can do this via translation:
Besides degenerate cases of the above like line segments, one might suspect these are the only polygons with this property. Indeed, in the general case it takes $\lceil\frac n2\rceil$ congruent $n$-gons to form an intersection similar to the original, so for $n>4$ we can't typically accomplish this. However, there are at least some exceptions:
In the above image, all non-right interior angles of the polygon are $135^\circ$, and the next three sides proceeding clockwise from the shortest side are $\sqrt{2}$ times as long as the previous. By making the turns in the "spiral" part of this construction shallower, one can find such polygons with arbitrarily many sides.
However, I believe such constructions approach an area ratio of $1$, and so one can't take a logarithmic spiral as a limiting case (because the intersection is no longer distinct from the original shapes).
I have two primary questions about such shapes:
Can we classify all polygons with this property?
Are there non-polygonal convex sets with this property?