Is there a (bounded) solid other than the sphere so that its orthogonal projection on any plane is always the same?
By "the same" I mean every projection is congruent to each other.
I believe the answer to this question is negative even if we weaken the notion of "same" to similar instead of congruent, but I see no easy way to prove either version.
Interestingly enough, the answer for the 2-dimensional version of the question is positive: any constant-width shape will do it. Of course, congruence among line segments is just a matter of length, so this is not very impressive. I wouldn't be surprised by the existence of many solids of constant shadow area.
Update. A solid whose orthogonal projections have the same area is known as a body of constant brightness and several examples do exist, contradicting Maxime Jaccon's comment this property is only held by the sphere.
Jaccon also mentioned such a solid should have constant width. I do believe this is true although I haven't manage to prove it. However, his follow up statement this property imply rotational invariance (a property only held by the sphere) is false as there are several solids of constant width, such as the revolution of constant width shapes and the Meissner's Tetrahedron.
According to Wikipedia, mathematician Soji Nakajima conjectured the sphere is the only solid of constant brightness and constant width. Furthermore, this has been proven in 2006, so the answer to my question should be, in fact, negative (and not an open problem). However, I still wonder how to prove this. It shouldn't be as hard as Nakajima's problem, which is a much stronger statement.
Clarification. As pointed out in the comments, if the convex hull of the solid is the sphere, then you cannot distinguish the projections of this solid from those of the sphere. That is not what I'm looking for, so when I ask for a solution "other than the sphere" I mean a solid with a distinct convex hull. The solid itself doesn't need to be convex, though.