Let $X$ and $Y$ be topological spaces and more precisely finite CW complexes. Let $f\colon X \to Y$ be a continuous map such that there exist a second continuous map $F\colon X^3 \to Y$ and $$ \forall x,y\in X: F(x,x,y)=F(x,y,x)=F(y,x,x)=F(x,x,x)=f(x) $$ Does that imply that $f$ is homotopic to a constant map?

Some partial results and motivation:

1. Consider a similar problem for groups with the usual product: If $X$ and $Y$ are groups $f\colon X \to Y$ and $F\colon X^3\to Y$ are group homomorphisms such that $F(x,x,y)=F(x,y,x)=F(y,x,x)=F(x,x,x)=f(x)$, then $$f(x)=F(x,x,x) = F(x,1,1)\cdot F(1,x,1) \cdot F(1,1,x) =(f(1))^3=1$$ and $f$ would be the zero map. This implies that $f$ induces a zero map in all homotopy groups.
2. If $X=Y$ and $f$ is the identity, the statement is true by the Whitehead theorem and the first point. This was already proven in 1977, Walter Taylor: Varieties obeying homotopy laws, Theorem 7.7
3. If $f$ is not the identity, it is not sufficent to be zero on all homotopy groups. For example the map from the torus to the sphere which contracts two circles induces a zero map in all homotopy groups, but is not homotopic to a constant map. That is my motivation: Finding a nice criteria for maps to be homotopic to a constant.
4. I have no counter example for more general $X$ and $Y$.
5. We can replace the space $Y$ by $X^3/M$ where $M$ is the equivalence relation identifying the majority. However, in general, $X^3/M$ is not contractible. It has for example nontrivial cohomology when $X$ is a circle.
6. If $X$ has dimension 1, the statement holds as this follows from the fundamental group.
7. If $X$ is a is a co-H-space in the category of pointed topological spaces, the set of maps $X\to Y$ up to homotopy becomes a group. In this case, also the statement holds. This includes the cases where $X$ is a suspension of another space. This result was brought to me by Prof. A. Thom.