Does $ f(pq)\times f((1-p)(1-q))=f(p(1-q))\times f(q(1-p))$ imply $f(pq)=f(p)\times f(q)$ over $[0,1]$?

Question

Let $f(x)$ be a non-negative Lebesgue measurable (or continuous, or differentiable, or strictly monotone, as needed) function defined on $[0,1]$.

Condition A: $f(x)$ satisfies $$ f(pq)\times f((1-p)(1-q))=f(p(1-q))\times f(q(1-p))$$ for all $p,q\in [0,1]$.

Condition B: $f(x)$ satisfies $$ f(pq)=k \times f(p)\times f(q)$$ where $k$ is a constant, for all $p,q\in [0,1]$.

Condition B obviously implies Condition A.

Question: Does Condition A imply Condition B?

Answer 1

No: $f(x)=2x$ satisfies Condition A but not Condition B.

Condition A is a homogeneous identity (both sides are the product of the same number of $f$-values), while Condition B is a non-homogeneous identity (the two sides are products of different numbers of $f$-values). Therefore even if one function $f$ satisfies both conditions, then any nontrivial multiple of $f$ will still satisfy Condition A but will no longer satisfy Condition B.

Edited to add: $f(x)=x^2$ also satisfies Condition A but not the new Condition B when $k\ne0,1$. Homogeneity also means that we can raise a solution to a power and get a new solution.