Supoose I have a closed curve $\gamma$ in the plane such that for any isometry $g$ of $\mathbb{E}^2,$ such that $g(\gamma)\neq \gamma,$ $\gamma$ intersects $g(\gamma)$ in at most two points. It should be an easy corollary of the fourvertex theorem that $\gamma$ is a circle, but I am not quite seeing it.
Assume $\gamma$ is smooth and it has two points $p_1$ and $p_2$ with different curvatures, say $\kappa_1>\kappa_2$. Then one can touch $\gamma$ at $p_2$ from inside by a $g(\gamma)$ such that $g(p_1)=p_2$. Since $\gamma$ and $g(\gamma)$ bound the same area, they intersect at some other points. (In fact at least 2, so together with $p_2$ it will be already 3.) By moving $g(\gamma)$ slightly, you can make at least 2 points of intersection near $p_2$, so all together it will be 3 points (or 4 if you read in the parenthesis). P.S. The same argument works if $\gamma$ is convex, but it require some Real analysis. 


In the case $\gamma$ is convex, there is an elementary argument which does not require smoothness. Denote by $\gamma_\alpha$ the rotation of $\gamma$ by an angle $\alpha$ around the center of mass of a region enclosed by $\gamma$. Since the areas and centers of mass of $\gamma$ and $\gamma_\alpha$ are equal, it is easy to see that they must have at least four points of intersection (if you don't see this immediately, see e.g. a one line proof in my book, Lemma 9.6). The only other possibility is $\gamma=\gamma_\alpha$. Since $\alpha$ can be arbitrary, we conclude that $\gamma$ is a circle. 


Igor Pak's argument takes care of the general convex case. So to complete the proof (without assuming any smoothness) it remains to consider the nonconvex case. To this end it is enough to note that if $\gamma$ is not convex, then it must intersect some line $L$ at least $4$ times. Let $g(\gamma)$ be the reflection of $\gamma$ with respect to $L$. Then $\gamma$ and $g(\gamma)$ intersect at least $4$ times. 

