= Solution
The function is holomorphic on a neighborhood of the closed unit disk. On $|z|=1$, the mapping assumption gives $|g(z)|<1=|z|$. Apply <Rouché's theorem> to $z$ and $-g(z)$. Their sum $z-g(z)$ has the same number of interior zeros as $z$, namely one counting <multiplicity>. Thus
$$
\boxed{g\text{ has exactly one fixed point in }|z|<1}.
$$
The zero is simple since its total <multiplicity> is one. No boundary <fixed point> is possible because $g$ maps the boundary strictly inside.
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