3 Sigh, another emendation; added 8 characters in body

I will break tradition of the accumulation of interesting comments and post a suggested partial solution (of something). The structure of mathoverflow is somehow not perfect for an incremental discussion (even though it is great in many ways).

The main idea so far for relevant contractive maps are folding the plane in half. So you could change the question and ask if any convex shape other than a circle has the property that it always fits inside of itself if you fold it once across a chord. [Edit 1: If you do ask, the answer is that one fold is not enough for a dented circle, as Martin points out in the comment.] A Rouleaux triangle does not have this property, but maybe it is interesting to check other regular Rouleaux polygons.

I think that no ellipse (other than a circle) has this property. You have to be a little careful because if you take an ellipse that is not round and not too thin, then if you fold it across its short semiaxis, the half-ellipse can fit inside of the original ellipse in non-standard ways.

If you fold the ellipse across its long semiaxis instead, then trivially the half ellipse only fits in the original ellipse in one way (up to symmetry). Suppose that you tilt this chord slightly, but keep it passing through the center, and then cut the ellipse in half. Then I think that this kind of half ellipse also fits in the original ellipse in one way. If that is correct, then if you fold the ellipse along this slightly tilted chord, then the folded shape does not fit in the original ellipse. [Edit 2: Anton says that it is not true, and that this half-ellipse which is cut at a slight diagonal can be movable within the original ellipse. I do not know whether it can be moved far enough, but I will refrain from speculating.]

A similar trick works for any regular odd-sided polygon $P$. $P$ has a longest diagonal. Make a chord which is close to this diagonal and parallel to it, so that the region on one side that has the majority of vertices has less than half of the area. This subregion only fits in $P$ in one way, so if you fold along this chord the folded shape does not fit. [Edit 2: At least this case of the argument actually works.]

I conjecture that if $K$ is any convex shape whose longest chords are isolated, then either by tilting or offsetting a longest chord, you can make a fold that does not fit in $K$. [Edit 2: A foolish conjecture as long as the ellipsoid case is in doubt.]

On the other hand, a constant-width body has the opposite property. It has an entire circle of longest chords, in a natural sense a maximal family of them.

2 Easy counterexample to one remark

I will break tradition of the accumulation of interesting comments and post a suggested partial solution (of something). The structure of mathoverflow is somehow not perfect for an incremental discussion (even though it is great in many ways).

The main idea so far for relevant contractive maps are folding the plane in half. So you could change the question and ask if any convex shape other than a circle has the property that it always fits inside of itself if you fold it once across a chord. [Edit: If you do ask, the answer is that one fold is not enough for a dented circle, as Martin points out in the comment.] A Rouleaux triangle does not have this property, but maybe it is interesting to check other regular Rouleaux polygons.

I think that no ellipse (other than a circle) has this property. You have to be a little careful because if you take an ellipse that is not round and not too thin, then if you fold it across its short semiaxis, the half-ellipse can fit inside of the original ellipse in non-standard ways.

If you fold the ellipse across its long semiaxis instead, then trivially the half ellipse only fits in the original ellipse in one way (up to symmetry). Suppose that you tilt this chord slightly, but keep it passing through the center, and then cut the ellipse in half. Then I think that this kind of half ellipse also fits in the original ellipse in one way. If that is correct, then if you fold the ellipse along this slightly tilted chord, then the folded shape does not fit in the original ellipse.

A similar trick works for any regular odd-sided polygon $P$. $P$ has a longest diagonal. Make a chord which is close to this diagonal and parallel to it, so that the region on one side that has the majority of vertices has less than half of the area. This subregion only fits in $P$ in one way, so if you fold along this chord the folded shape does not fit.

I conjecture that if $K$ is any convex shape whose longest chords are isolated, then either by tilting or offsetting a longest chord, you can make a fold that does not fit in $K$.

On the other hand, a constant-width body has the opposite property. It has an entire circle of longest chords, in a natural sense a maximal family of them.

1

I will break tradition of the accumulation of interesting comments and post a suggested partial solution (of something). The structure of mathoverflow is somehow not perfect for an incremental discussion (even though it is great in many ways).

The main idea so far for relevant contractive maps are folding the plane in half. So you could change the question and ask if any convex shape other than a circle has the property that it always fits inside of itself if you fold it once across a chord. A Rouleaux triangle does not have this property, but maybe it is interesting to check other regular Rouleaux polygons.

I think that no ellipse (other than a circle) has this property. You have to be a little careful because if you take an ellipse that is not round and not too thin, then if you fold it across its short semiaxis, the half-ellipse can fit inside of the original ellipse in non-standard ways.

If you fold the ellipse across its long semiaxis instead, then trivially the half ellipse only fits in the original ellipse in one way (up to symmetry). Suppose that you tilt this chord slightly, but keep it passing through the center, and then cut the ellipse in half. Then I think that this kind of half ellipse also fits in the original ellipse in one way. If that is correct, then if you fold the ellipse along this slightly tilted chord, then the folded shape does not fit in the original ellipse.

A similar trick works for any regular odd-sided polygon $P$. $P$ has a longest diagonal. Make a chord which is close to this diagonal and parallel to it, so that the region on one side that has the majority of vertices has less than half of the area. This subregion only fits in $P$ in one way, so if you fold along this chord the folded shape does not fit.

I conjecture that if $K$ is any convex shape whose longest chords are isolated, then either by tilting or offsetting a longest chord, you can make a fold that does not fit in $K$.

On the other hand, a constant-width body has the opposite property. It has an entire circle of longest chords, in a natural sense a maximal family of them.