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?

8If I understand well, you don't ask whether $f$ and $g$ have a common fixed point. Yet, the answers given so far speak of fixed points ... – Denis Serre – 2011-03-31T14:28:29.460

2Could more be true? Would f and g necessarily have a common fixed point? This might perhaps be easier to prove if true. – Harald Hanche-Olsen – 2009-10-29T23:35:35.097

9Apparently f and g may not have a common fixed point even in the dimension 1 case. This is mentioned in the first paragraph of "Equivalent Conditions involving Common Fixed Points in the Unit Interval" by Jachymski. Unfortunately I can't follow the reference given. @fedja, what an amazing problem! – Alon Amit – 2009-10-30T16:58:33.860

7

Ah, indeed: Jachymski refers to an abstract in an old Notices. But searching MR reveals J.P. Huneke: On common fixed points of commuting continuous functions on an interval. Trans. Amer. Math. Soc. 139 1969 371--381. See http://www.jstor.org/stable/1995330 if you have JSTOR access. From the abstract: “This paper offers two methods of constructing commuting pairs of continuous functions [...] which map [0,1] to itself without common fixed points”. Jachymski also notes that if the iterates of one function forms an equicontinuous family, there is a common fixed point.

– Harald Hanche-Olsen – 2009-10-30T18:02:53.127A paper of Christian Bonatti is vaguely related: "Un point fixe commun pour des difféomorphismes commutants de S^2", Annals of Maths 1989. – Benoît Kloeckner – 2014-01-10T20:14:43.423