Question

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 four-vertex theorem that $\gamma$ is a circle, but I am not quite seeing it.

Answer 1

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.

Answer 2

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.
If $\gamma$ is not convex then it has concave and convex points. In this case you can touch a concave point from inside by a convex one and the same proof works.