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3 deleted 164 characters in body

we can clearly assume that $f(z)=-z$ in the standard metric on $S^1\subset \mathbb C$ (as we can assume that $f$ is isometric with respect to some Riemannnian metric on $S^1$). Then $g(z)=z\cdot e^{i\alpha(z)}$ with $0<\alpha(z)<2\pi$. If $\alpha(z_0)<\pi$ then $\alpha(g(z_0))>\pi$ and there is a point $z_1$ with $\alpha(z_1)=\pi$ by the intermediate value theorem.

BTW, could you please change $\mathbb Z^2$ to $\mathbb Z_2$ in the question? I find the notation $\mathbb Z^2$ for a cyclic group of order $2$ rather disconcerting.

Post Undeleted by Vitali Kapovitch
2 added 85 characters in body

seems pretty obvious to me.

we can clearly assume that $f(x)=-x$ f(z)=-z$in the standard metric on$S^1\subset \mathbb C$(as we can assume that$f$is isometric with respect to some riemannnian Riemannnian metric on$S^1$). Then$g$must have g(z)=z\cdot e^{i\alpha(z)}$ with $0<\alpha(z)<2\pi$. If $\alpha(z_0)<\pi$ then $\alpha(g(z_0))>\pi$ and there is a point $x_0$ where $g(x_0)=-x_0$ since otherwise z_1$with$deg(g)=0$. \alpha(z_1)=\pi$ by the intermediate value theorem.

BTW, could you please change $\mathbb Z^2$ to $\mathbb Z_2$ in the question? I find the notation $\mathbb Z^2$ for a cyclic group of order $2$ rather disconcerting.

Post Deleted by Vitali Kapovitch
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