Let $f \ge 1$ be a multiplicative arithmetic function and $F_f(n) = F(n)= \sum_{d|n}f(d)$. Define the entropy of $n$ with respect to $f$ to be

$H_f(n) = -\sum_{d|n} \frac{f(d)}{F(n)}\log(\frac{f(d)}{F(n)}) = \log(F(n)) - \frac{1}{F(n)}\sum_{d|n} f(d)\log(f(d))$

If $E_f(n) = \sum_{d|n} f(d)\log(f(d))$ then using Möbuis inversion one can show that from $E_f = E_g$ it follows that $f=g$. Similarily from $F_f = F_g$ it follows that $f=g$. ( As a byside: The function $E_f$ satisfies a Leibniz rule: $E_f(nm) = F_f(m)E_f(n)+E_f(m)F_f(n)$ for $m,n$ with $\gcd(m,n) = 1$.)

Now my question is: Does it from $H_f = H_g$ necessarily follow that $f=g$, or is there a counterexample?