Is circle the only shape that can remain convex after folding? Here "fold" means "fold a piece of paper (along a straight crease)". The sketch below shows that one can always find a fold by which an ellipse or rectangle loses convexity. But it seems a circle remains convex no matter how the crease is chosen?
I am not sure how to deal with a problem like this, where the shape is generic and a convenient coordinate system cannot be defined.
Update: to be precise, the circle means "disk", which includes both the border and the interior. Same for all other shapes.

Update Jan. 13 I am thrilled by the number of upvotes. Here is an interesting experiment. Starting simple, let's use an upright rectangle centered at the origin, and reflect it w.r.t. all creases passing the origin (the experiment can only use a finite subset of such creases). The area union of all reflected shapes appears to be a disk.

This observation seems to allow the following general statement: a 2d shape, convex or not, transforms into a new shape by folding. The union of all new shapes is a disk.
To be honest, this statement itself sounds like a question in need of a proof. But I feel it is highly related with the original question. So I update it here along with the Mathematica code for plotting the figure (). Enjoy!
foldAlongMiddleAxis[mya_, mytheta_] := 
 Block[{a = mya, cs = Cos[mytheta], 
   ss = Sin[mytheta]},
  polPrime1 = 
   Polygon[{{cs, ss}, {cs, ss} + a {-ss, cs}, {-cs, -ss} + 
      a {-ss, cs}, {-cs, -ss}}];
  polPrime2 = Polygon[{{1, 0}, {1, a}, {-1, a}, {-1, 0}}]; 
  Graphics[{{Opacity[0], EdgeForm[Gray], polPrime2}, {Opacity[0], 
     EdgeForm[Gray], polPrime1}}, AspectRatio -> Automatic]]

plotAllFoldsAndTrajO[a_, n_] := Show[{
   Graphics[{LightPink, Opacity[0.9], EdgeForm[Gray], 
     Polygon[{{1, -a}, {1, a}, {-1, a}, {-1, -a}}]}],
   Table[foldAlongMiddleAxis[a, mytheta], {mytheta, 
     Range[0, 2 \[Pi], \[Pi]/n]}],
   ParametricPlot[{{Cos[\[Theta]], Sin[\[Theta]]} + 
      a {-Sin[\[Theta]], Cos[\[Theta]]}}, {\[Theta], 0, 2 \[Pi]}, 
    PlotStyle -> {Black, Thick}]
   }]

plotAllFoldsAndTrajO[5, 50]

 A: Sketch Proof
The boundary of the original convex shape is differentiable almost everywhere. Let $A,B,C$ be three points on the boundary where the boundary is differentiable and let $t_A,t_B,t_C$ be the respective tangents.
Folding along the perpendicular bisector of $AB$ shows that $t_A$ and $t_B$ must be equally but oppositely inclined to $AB$. Similarly for $t_B$ and $t_C$ and for $t_C$ and $t_A$. Then $t_A,t_B,t_C$ are tangents to the circle through $A,B,C$.
This circle is therefore determined by $A,B$ alone and so contains every point on the boundary where the boundary is differentiable. It therefore contains every point on the boundary.
A: This isn't a complete answer, more like a broad sketch of a proof, but it's too long for a comment.
First, we can see that the region must be convex, since if it is not convex, a very tiny fold (think of the fold as being a line segment joining two points on the boundary of the region, so if a fold is "tiny" the length of that segment is very small) will not appreciably alter the shape, thus it will remain non-convex.
Second, it is intuitive (but no proof) that a convex polygon cannot work, because in such a shape, there must be an interior angle where two adjacent sides meet and has measure less than $\pi$.  Then since a polygon must have at least 3 sides, there is at least one other side not adjacent to this angle, which must form a triangle with the other two sides (the angle cannot point "the wrong way" because the region is convex).  Therefore, there is a line of reflection between this side and the vertex of the subtending angle that reflects the vertex across to the other side, and this results in a non-convex shape.
Then we can generalize further to exclude any convex shapes that have a polygonal vertex, since a similar argument to the above shows that there must be a line that will reflect that vertex in a way that will make it "poke out" of any smooth curve.
So we know the region must be smooth.  Then I believe the remaining argument is to show that the boundary must have constant curvature.  This strikes me as a previously solved problem but I don't have a literature reference.
