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