First by the Brouwer fixed point theorem $f$ has at least one fixed point, say $\bar{x}=f(\bar{x})$.
If that fixed point is unique (contraction mappings spring to mind for a bunch of examples of this) we're done since $g(\bar{x})=g(f(\bar{x})=f(g(\bar{x}))$ g(\bar{x})=g(f(\bar{x}))=f(g(\bar{x}))$and we see that$g(\bar{x})$is "another" fixed point of$f$, since the fixed point was unique$g(\bar{x})=\bar{x}=f(\bar{x})$. For "less nice"$f$we still have that$f(g(\bar{x}))=g(\bar{x})$... in fact for any$n\in\mathbb{N}$we have$f(g^n(\bar{x}))=g^n(\bar{x})$. If (without resorting to sequences)$g^n(\bar{x})\to y$, we can again claim success since we'll have$g(y)=y$and$f(y)=y$. Unless there's another "obvious" easy case I missed it seems like the interesting cases will be when$g^n(\bar{x})$does not converge. Two sub-cases spring to mind: when$g^{n}(\bar{x})$has finitely many accumulation points (like when$g^n(\bar{x})$is a periodic point of$g$), or ... it has lots. Intuition (really thinking about rational and then irrational rotations about the origin as one way to generate those two cases) tells me that in either of these cases what we really need to do drop the$\bar{x}$as a "starting point". It "would be nice" if we can show$g$conjugate to a rotation in the above two cases. Any thought on if that is true or not? I suspect not else$g$would have a unique fixed point and we'd be done (as above)... Maybe semi-conjugate... but would that help? New minds, any thoughts? 1 I'm sure what I write has been thought of by many, but it's a starting point that I thought should be written down. First by the Brouwer fixed point theorem$f$has at least one fixed point, say$\bar{x}=f(\bar{x})$. If that fixed point is unique (contraction mappings spring to mind for a bunch of examples of this) we're done since$g(\bar{x})=g(f(\bar{x})=f(g(\bar{x}))$and we see that$g(\bar{x})$is "another" fixed point of$f$, since the fixed point was unique$g(\bar{x})=\bar{x}=f(\bar{x})$. For "less nice"$f$we still have that$f(g(\bar{x}))=g(\bar{x})$... in fact for any$n\in\mathbb{N}$we have$f(g^n(\bar{x}))=g^n(\bar{x})$. If (without resorting to sequences)$g^n(\bar{x})\to y$, we can again claim success since we'll have$g(y)=y$and$f(y)=y$. Unless there's another "obvious" easy case I missed it seems like the interesting cases will be when$g^n(\bar{x})$does not converge. Two sub-cases spring to mind: when$g^{n}(\bar{x})$has finitely many accumulation points (like when$g^n(\bar{x})$is a periodic point of$g$), or ... it has lots. Intuition (really thinking about rational and then irrational rotations about the origin as one way to generate those two cases) tells me that in either of these cases what we really need to do drop the$\bar{x}$as a "starting point". It "would be nice" if we can show$g$conjugate to a rotation in the above two cases. Any thought on if that is true or not? I suspect not else$g\$ would have a unique fixed point and we'd be done (as above)... Maybe semi-conjugate... but would that help? New minds, any thoughts?