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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 . (In fact at least 2. , so together with $p_2$ it will be already 3.) By moving $g(\gamma)$ slightly, you can make 4 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.

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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. By moving $g(\gamma)$ slightly, you can make 4 points of intersection.