Prove that there are maximum 4 points on an arbitrary given closed convex Jordan curve of $C$ on a plane which for each one like $p$ we have the following property:
If we rotate $C$ around $p$ by $\frac {\pi} {2}$ clockwise on the plane then the created curve $C'$ has no intersection with $C$ exept on $p$.
What if we change $90^{\circ}$ by another degree or change $C$ to a non convex closed jordan curve?
Update:This proof is only valid for convex curves.
First consider $P$ and $Q$ are two points on $C$ satisfying the condition ,then there are two angles $\angle lPl'$ = $\angle hQh'$ = $90^{\circ}$ which $P$ is inside $\angle hQh'$ and $Q$ is inside $\angle lPl'$. (Why?)
let $A$ intersection of the semilines $l$ ,$h$ and $B$ intersection of semilines $l'$ ,$h'$ ,now we can easily see that $C$ is inscribed in convex quadrilateral $APBQ$.(why?)
there are two options now:
1.one of two angles $\widehat{PAQ}$ ,$\widehat{PBQ}$ are less than $90^{\circ}$ and the other is bigger.
let $\widehat{PAQ}$ < $90^{\circ}$ and $\widehat{PBQ}$ > $90^{\circ}$. now we can see for every point $b \in C \not= P,Q$ inside or on the triangle $\triangle PBQ$ does not have the condition because $\widehat{PbQ}$ > $\widehat{PBQ}$ > $90^{\circ}$ and $P,Q \in C$.
so if there are another points which have the condition ,they must be inside or on the $\triangle PAQ$ .we consider another point having the condition $R \in C$ among all $C$ points inside or on the triangle $\triangle PAQ$ which the angle $\widehat{PRQ}$ < $90^{\circ}$ be the minimum. we can easily investigate there's no point other than $R$ having such condition inside or on the triangle $\triangle PAQ$ and hence in the whole quadrilateral $APBQ$ which contains $C$.(how?)
so we find maximum 3 point having the condition in this option.
2.$\widehat{PAQ} = \widehat{PBQ} = 90^{\circ}$
in this option we can have only maximum two point have the condition and they must be $A$ and $B$.(why?)
so we find maximum 4 points on $C$ having the property which are edges of a rectangle containing $C$.