= Solution
\b[Yes: a tangential zero and a crossing zero can give exactly two <fixed points>.] One explicit <continuous> piecewise linear example is
$$
\boxed{f(x)=\begin{cases}
1-2x,&0\le x\le1/3,\\
\frac52x-\frac12,&1/3\le x\le1/2,\\
1-\frac12x,&1/2\le x\le2/3,\\
2-2x,&2/3\le x\le1.
\end{cases}}
$$
Adjacent formulas agree at each junction. Their endpoint values lie in $[0,1]$, so linear interpolation proves that the whole image lies in $[0,1]$; also $f(0)=1$ and $f(1)=0$. On the four pieces, $f(x)-x$ is respectively $1-3x$, $3x/2-1/2$, $1-3x/2$, and $2-3x$. Their only zeros on their respective intervals are $1/3$ and $2/3$. Thus \b[the complete <fixed point> set is $\boxed{\{1/3,2/3\}}$]. Continuity forces at least one zero, but does not force every zero to change sign.
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