A question on continuous maps from Möbius to itself

Question

Let $M$ denotes the Möbius strip. Then is it true that

For every continuous map $f:M\to M$ there is $x\in M^\circ$ ($x\notin\partial M$) such that $f(f(x))=x$?

Answer 1

I might be missing something, but for $f = g\circ h,$ where $h$ is the retraction onto the core circle (see this question), and $g$ is the rotation of the circle by, say, 1 radian there appears to be no such $x.$

Answer 2

No. Think of the Moebius strip as a quotient of the trivial bundle over the circle by an involution $(x,y)\mapsto(x+1,-y)$. Give it the quotient Riemannian metric by the involution, from the product metric, so $dx^2+dy^2$. Take the usual vector field of rotation of the circle. Lift to a vector field on trivial bundle, perpendicular to the fibers, and quotient to the Mobius strip. The nonzero vector field has well defined flow, since its lift to the trivial bundle is complete. The vector field is nowhere zero, so the flow has no fixed points.