I don't know the answers. Consider this as an extended comment.
Let $P$ be the set of fixed points of $g$. Then $f$ maps $P$ into itself.
Indeed let $x\in P$, so $g(x)=x$. Then
$$f(x)=f(g(x))=g(f(x)),$$
which means that $f(x)\in P$. Now $P$ is finite, so $f$ must have
a periodic point in $P$. So $f^m$ and $g$ have a common fixed point.
Of course one can interchange $f$ and $g$ here.

With little more care, one can find a common fixed point of $f^m$ and $g^n$
with some $m,n$, which is REPELLING for both $F=f^m$ and $G=g^n$.
Let this common repelling fixed point be $a$.

Evidently $F$ commutes with $G$. If two communing functions share a
repelling fixed
point, then they have the same Poincare function at this point.
Poincare function $\phi$ is the ``linearizer'', that is the solution
of the functional equation
$$\phi(\lambda z)=F(\phi(z)), \quad \phi(0)=a,\quad\phi'(0)=1,
$$
where $\lambda=F'(a)$.

Now if $F'(a)=G'(a)$ we easily conclude that $F=G$.
Which means that $f$ and $g$ have a common iterate.
If $F'(a)\neq G'(a)$ there is only a very restricted set of possibilities
which were completely described and classified by Ritt.

Thus we come to a question:

*Suppose that $f$ and $g$ have a common iterate.
Must they have a common fixed point?*

Or, perhaps those pairs which do not
permit some explicit classification?

The question about rational functions that have a common iterate is interesting in itself
and I am posting it separately, Rational functions with a common iterate