 Solution

Yes: a tangential zero and a crossing zero can give exactly two fixed points. One explicit continuous piecewise linear example isAdjacent 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 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.