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Suppose $X$ is a metric space and $f,g$ are self maps on $X$. If $f$ is a cntracton and $f(g(x)) = g(f(x)) $, does $g$ have a fixed point?

I know $f$ has a unique fixed point as a consequence of contraction but I am stuck.

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    $X$ is required to be complete. As a counterexample, consider $X=(0,1]$ and the map $f(x)=\frac{x}{2}$. Then $f$ is a contraction, but has no fixed point. – 2017-01-26
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    If $g(x)=x$ then $g$ has much more fix points than $f$. – 2017-01-27

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Let $c$ be the unique fixed point of $f$.

Then $f(g(c)) = g(f(c)) = g(c)$. Thus, $g(c)$ is a fixed point of $f$.

By uniqueness, we get $g(c) = c$.

But ...

The above argument only works provided you know $f$ has a fixed point.

As Aweygan notes, without a stronger hypothesis, the fact that $f$ is contractive doesn't allow you to assert that $f$ has a fixed point.

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    Assuming $X$ is complete. The answer still has issues. There's still no guarantee $g$ has a unique fixed point. – 2017-01-26
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    The problem was to show that $g$ has a fixed point. There was no requirement to show uniqeness of $g$'s fixed point. If $X$ is complete, then $f$ has a unique fixed point, in which case, the proof I posted shows that $g$ has at least one fixed point. – 2017-01-26