Are pullbacks on singular cohomology unique on the nose? Let $C$ be the category of even-dimensional connected closed oriented topological manifolds/orientation-preserving continuous maps and $D$ be the category of finite-dimensional graded $\mathbb{Q}$-algebras. 
We have a functor $H:C\rightarrow D$ given by singular cohomology. 
Let us call a choice of an isomorphism of graded $\mathbb{Q}$-algebras $H(M)\otimes H(M')\rightarrow H(M\times M')$ for every $M$, $M'\in Obj(C)$ a Kunneth system. Projections $M\times M'\rightarrow M$, $M'$ define a Kunneth system which is functorial with respect to $H$ (Kunneth theorem). 
For a contravariant functor $G:C\rightarrow D$ such that the induced map on objects is equal to one induced by $H$, it is meaningful to ask if the Kunneth system above is functorial with respect to $G$. The question: does there exist a contravariant functor $G:C\rightarrow D$ such that 


*

*$G$ coincides with $H$ on the level of objects but not on the level of morphisms;

*for any non-negative even integer $i$, the contravariant functors from $C$ to $\mathbb{Q}$-vector spaces obtained by composing either $G$ or $H$ with the "$i$-th graded piece" functor are equal; 

*the Kunneth system above is functorial with respect to $G$? 


P.S.: this question is inspired by this question. 
 A: The answer to the question as posed is no, if I have understood the setup correctly.
It suffices to show that a contravariant functor $G:C\to D$ satisfying your second and third bullet points must agree with $H:C\to D$ on morphisms. This would follow if we can show that for every $a\in H^i(M)=G^i(M)$ with $i$ odd and for every morphism $f:N\to M$ in $C$ we have $H(f)(a)=G(f)(a) \in H^i(N)=G^i(N)$. 
To see this, take a second class $b\in H^j(M')=G^j(M')$ with $j$ odd, and a second morphism $g:N'\to M'$. Observe that by naturality of the Kunneth isomorphisms for $G$ and $H$ and the fact that $a\times b\in H^{i+j}(M\times M')$ with $i+j$ even, so that $H(f\times g)(a\times b)=G(f\times g)(a\times b)$, we have that 
$$G(f)(a)\otimes G(g)(b)=H(f)(a)\otimes H(g)(b) \in H^i(N)\otimes H^j(N') = G^i(N)\otimes G^j(N').$$
The tensor product is over the rationals. We may take $N'=M'$ and $g=\operatorname{Id}_{M'}$, the identity map. By functorialty $G(g)(b)=b=H(g)(b)$, and so 
$$
G(f)(a)\otimes b=H(f)(a)\otimes b
$$
which by bilinearity of tensor products gives that
$$(G(f)(a)-H(f)(a))\otimes b = 0.$$
Choosing $b$ to be nonzero, this implies that $G(f)(a)=H(f)(a)$, as claimed. 
