NOTE After posting this question, I realised I had not defined the overall shape of the form. The final note at the bottom was intended to correct that omission. I am no longer certain how this affects things. On the face of it, it defeats my argument (see answer by Ethan Bolker). Nevertheless, I'll let the question stand (see comment by Joshua Wang under EB's answer).
I have looked for a previous thread on this (including under Mobius strip) and have not found the precise same question.
I am not a mathematician so please forgive any incorrect terminology.
I'll define a bracelet as a flat ribbon or band of metal, joined to itself end-to-end with any number of half-twists. Assume that during manufacture the twisting is made uniform along the length of the bracelet.
I'm interested in the wheel-like rotational symmetry about the axis and the placement of the centre of gravity. Intuitively it seems obvious to me that if twisting is uniform, all bracelets have a balance point at the centre** except for the one with a single half-twist.
How can I prove (a) that the C of G is non-central for the single-twist example and (b) that the C of G is central for all other numbers of twists whether zero, odd or even.***
*** Negative twists are not considered distinct for these purposes.
** Let us say that the bracelet is formed in such a way that a central line drawn within and along the length of its volume forms a perfect circle. The centre is then the centre of this circle.
Are you sure the center of gravity is not the central point in all cases? For a single half twist the bracelet is vertical (that is, parallel to the axis) at one end of one of the diagonals and horizontal at the other end with the center line on the circle. The moments will be the same.
I think this kind of argument shows that the center of gravity will always be at the center of the circle you describe.