These ($f=c>0$ on $(a,b)$ with $b-a=1$) are the only examples. Write $m=\min f$, $M=\max f$. Then a solution to $f(p)=\|f\|_p$ would have to satisfy
$$
m(b-a)^{1/x}\le f(x)\le M(b-a)^{1/x}, \quad\quad\quad (1)
$$
and since $x$ can be a point where the min/max is assumed, this rules out $b-a\not=1$. If $b-a=1$ and $f$ is not constant, then we obtain strict inequalities in (1) and the same argument still works.