Is there a dense subset on closed Jordan curve $C$ which its points make intersections under certain rotations? Is it true that for any given closed Jordan curve of $C \subset \mathbb{R}^2$ there is a dense subset $A$ such that for every point $p\in A$ we have the following property:
If we rotate $C$ around $p$ by 0 < $\alpha$ < $\pi$ clockwise on the plane, then the created curve $C'$ intersects $C$ at points other than $p$.
What if the angle of rotation becomes $\pi$ ?
 A: The question as stated still does not make much sense, and I think, this is why it does not attract the attention it deserves. Let me restate the problem and tell what I have so far.

For a Jordan curve $C$ on the plane $\Pi$ and some angle $\alpha\in (-\pi,\pi]\backslash\{0\}$ the set $R(\alpha)$ consists of all points $p$ with the following property: if we rotate $C$ around $p$ by $\alpha$ the resulting curve $C'$ intersects $C$ precisely at the point $p$. What can be said about $R(\alpha)$?

Let me start with some topology. By Jordan's theorem $C$ separates $\Pi$ into a bounded and unbounded components with $C$ being their common boundary. Let $B$ be the open bounded component, and $D$ be its closure, i.e. $B=D\sqcup C$. Define $B'$ and $D'$ analogously and assume that $C\cap C'=\{p\}$. 

We have $D\cap D'=\{p\}$.

Fist, assume that there are $q\in C'\cap B$, and $r\in C'\backslash D$. Then $q,r$ separate the Jordan curve $C'$ into two arcs, which join $q,r$ and intersect only in $q,r$. Since $q,r$ are in different components of $\Pi\backslash C$, both of the arcs must intersect $C$, and so contain points of $C\cap C'=\{p\}$. Hence, either $q=p$ or $r=p$, which is impossible, since $q,r\not\in C$. Consequently, we either have that $C'\subset D$, or $C'\cap B=\varnothing$. Assume that $C'\subset D$. Then it is easy to see that $D'\subset D$, and so, $D'$ is strictly contained in $D$, which is impossible, as they are isometric.
Therefore, $B\subset \Pi\backslash C'=B'\sqcup(\Pi\backslash D')$. Since $B$ is connected, either $B\subset B'$, or $B\subset \Pi\backslash D'$. The first option is again impossible, since it implies that $D$ is strictly contained in $D'$. Thus, $B\subset \Pi\backslash D'$ and analogously $B'\subset \Pi\backslash D$, from which the claim follows.

Now consider the smooth case. Namely:

If $p\in R(\alpha)$ is a point of smoothness for $C$, then $\alpha=\pi$.

Indeed, assume that in some local coordinates $C$ and $C'$ can be represented as $y=f(x)$ and $y=g(x)$, $x\in(-1,1)$, with $p=(0,0)$. Let $G=\{(x,f(x))|x\in (-1,1)\}$, which is homeomorphic to an open interval. Then $C\backslash G$ is compact and not containing $p$, and so there is $\varepsilon>0$ such that the set $Q=(-\varepsilon,\varepsilon)\times (-\varepsilon,\varepsilon)$ does not intersect $C\backslash G$. Define $Q_+=\{(x,y)\in Q|y>f(x)\}$ and $Q_-=\{(x,y)\in Q|y<f(x)\}$, which are both connected and not intersecting $C$. Hence, for each of $Q_+$ and $Q_-$ there is a component of $\Pi\backslash C$ which contains it. Moreover, these components are different. Indeed, if say both $Q_{\pm}\subset B$, then $B(p,\varepsilon)\subset D$, and so $p$ is not in the closure of $\Pi\backslash D$. Contradiction.
Now, assume that $\alpha\ne\pi$. Then the tangent lines at $0$ are not parallel, and so $f'(0)\ne g'(0)$. WLOG we can assume that $h_0=h'(0)>0$, where $h=f-g$. Then there is $\delta>0$, such that $\left|\frac{h(x)}{x}-h_0\right|<h_0$, when $x\in (-\delta,0)\cup (0,\delta)$, and in particular $h(x)<0$, when $x\in (-\delta,0)$ and $h(x)>0$, when $x\in (0,\delta)$. Hence, $f(x)<g(x)$, when $x\in (-\delta,0)$ and $f(x)>g(x)$, when $x\in (0,\delta)$. Thus, if $0<x<\min\{\varepsilon,\delta\}$, then $(x,g(x))\in Q_-$, and if $0>x>\min\{\varepsilon,\delta\}$, then $(x,g(x))\in Q_+$. Since either $Q_-$ or $Q_+$ are contained in $B$ we conclude that there is a point in $C'\cap B$ other than $p$. Contradiction.
We get the following conclusion:

If $C$ is smooth and $\alpha\in (-\pi,\pi)\backslash\{0\}$, then $R(\alpha)=\varnothing$.

Hence, the problem for the smooth curves is not very interesting.

Note that almost every point of $C$ is a point of smoothness when $C$ is a convex curve (i.e. $D$ is a convex set), but in this case much more can be said. First,

If $C$ is a strictly convex curve, then $C=R(\pi)$. Conversely, if $R(\pi)$ is dense in $C$, then $C$ is strictly convex.

Recall that if $D$ is a convex set and $p\in C=\partial D$, there is a support line - a line $l$ passing through $p$ such that $l\cap D\subset C$. If $D$ is strictly convex, then $l\cap D=\{p\}$. Since $D$ lies in only one of the half-planes with respect to $l$, the rotation in $p$ by $\pi$ (i.e. the reflection with respect to $p$) moves $D$ into a set $D'$ in a different half-plane, and so $D\cap D'\subset l$, which implies $D\cap D'=\{p\}$. Thus, $p\in R(\pi)$.
Conversely, assume that $R(\pi)$ is dense in $C$. First, note that the map $x,y\to\frac{x+y}{2}$ is continuous and open. Since $B$ is open and connected it follows that $\frac{1}{2}B+\frac{1}{2}B$ is also open and connected. Moreover, $(\frac{1}{2}B+\frac{1}{2}B)\cap R(\pi)=\varnothing$. Indeed, if $x,y\in B$ we have that the reflection in $\frac{x+y}{2}$ moves $x\in B$ into $y\in B$, which would mean that $y\in B\cap B'$, which is impossible.
Since $\Pi\backslash(\frac{1}{2}B+\frac{1}{2}B)$ is closed and it contains $R(\pi)$, it also contains its closure, which is $C$. Hence, $\frac{1}{2}B+\frac{1}{2}B$ is connected, intersecting $B$ and not intersecting $C$. Hence, $\frac{1}{2}B+\frac{1}{2}B\subset B$, and again, from the continuity we have that $\frac{1}{2}D+\frac{1}{2}D\subset D$, and so for every $x,y\in D$ and every dyadic rational $\mu\in [0,1]$ we have $\mu x+ (1-\mu)y\in D$. Since such points are dense in the segment joining $x$ and $y$, and $D$ is closed, we see that the whole segment is contained in $D$. Since $x,y$ are arbitrary we conclude that $D$ is convex.
Finally, $C$ cannot contain a straight segment, since if $[x,y]\subset C$, then the reflection in $\frac{x+y}{2}$ moves $x\in C$ into $y\in C$, which would mean that $y\in C\cap C'$, which is impossible. Thus, $D$ is strictly convex.

If $C$ is convex and $p\in R(\alpha)$, for $0<|\alpha|<\pi$, then there are support lines to $D$ at $p$ such that the angle between them is less or equal to $|\alpha|$. Conversely, if such support lines exist for some $\alpha>0$, then $p\in R(\beta)$, for any $\pi\ge|\beta|>\alpha$.

This is more or less obvious, since if $p\in R(\alpha)$, as $D$ is convex, for any $q,r\in C$ the solid triangle $\Delta pqr \subset D$. Its image under rotation by $\alpha$ is contained in $D'$ and so can only intersect with $D$ in $p$. In particular, the two triangles only intersect in $p$, which is only possible when $\angle qpr \le |\alpha|$. If the two support lines do exist, then $D$ is completely confined within two rays radiating from $p$ with angle $\alpha$. Then any rotation by more than $\alpha$ moves this region away from itself.

If $C$ is convex and $\alpha\in (0,\pi)$ then $\bigcup_{0<\beta\le \alpha}R(\beta)$ contains at most $\frac{2\pi}{\pi-\alpha}$ points. Analogous estimate is true for $\alpha\in (-\pi,0)$. (This is mostly inspired by Solar Galaxy's answer)

Let $p_i\in R(\beta_i)$, for $i\in \overline{1,n}$ and $0<\beta_i\le \alpha$, and assume that they are numerated in this particular order clockwise. Then $\angle p_{i-1} p_i p_{i+1} \le\beta_i\le \alpha$, for every $i\in \overline{1,n}$ ($\pm 1$ is implemented $\mod n$). Recall, that the sum of the angles of the polygon with $n$ vertices is $(n-2)\pi$. Hence, $(n-2)\pi\le n\alpha$, from which the estimate follows.

Hence, if $C$ is convex, then $R(\alpha)$ is finite, and moreover $\bigcup_{-\pi<\alpha<\pi }R(\alpha)$ is at most countable. I don't think more can be said: there are convex curves, which are not smooth in a dense set of points, and then $\bigcup_{-\pi<\alpha<0,0<\alpha<\pi }R(\alpha)$ will be dense (albeit countable).
So, one can try to show that if $R(\alpha)=C$, for some $\alpha$ it also implies convexity, and we obtain a contradiction immediately, but I guess, that is a difficult part.
Also this problems admits a generalization that I've posted earlier, and which I cannot solve even in the smooth case, because the difficulty here lies in the global geometry, not local.
Enveloping a Jordan curve with a trace of another one
A: For convex closed curves and for $\frac {\pi} {2}$ rotation I have found maximum 4 point on every $C$ which has no intersection under the following rotation ,also for $\frac {\pi} {3}\le\alpha$ < $\frac {\pi} {2}$ maximum 3 points and for $\alpha$ < $\frac {\pi} {3}$ there must be maximum 2 points satisfying the condition, I'm investigating for $\frac {\pi} {2}$ < $\alpha$...
https://math.stackexchange.com/questions/2773404/prove-that-there-are-maximum-4-points-on-every-closed-jordan-curve-c-which-avo
A: Thanks for the efforts have done on this problem specially by @erz  ,but It seems that Mark J.Nielsen have solved this problem here (that I have found it recently), while he was proving this theorem about inscribed triangles in closed simple jordan curves:
Theorem E: Extending Theorem D, J has so many inscribed triangles similar to T that the vertices of all these inscribed triangles are "dense" in the curve J.
Actually the statement of this theroem is equal to the problem above ,therefore the answer of this problem is affirmative.
As the article needs license I do not have access to the whole solution ,I think it would be great to present and discuss about the way the problem has been solved Here.
