Do Steenrod Squares have naturality with homomorphisms that don't come from continous maps

Question

I was reading about the Steenrod Squaring operations in Milnor and Stasheff's, Characteristic Classes, now there is an axiom regarding naturality that says that given a continous map $f:(X,Y)\rightarrow (X',Y')$ and $f^*$ is the induced map on cohomology groups then $Sq^i\circ f^*=f^*\circ Sq^i$.
Given a map $g^*$ between cohomology groups that does not come from a continous map between topological spaces do we still get the naturality with the Steenrod Squares ?
The motivation behind this is that, assuming coefficients in $\mathbb{Z}_2$, there is a monomorphism from the cohomology of the real grassmannian to the cohomology of $\mathbb{R}P^\infty\times ...\times\mathbb{R} P^\infty$, as the grassmannian is a universal bundle and the squaring operations are easily computable on $\mathbb{R}P^\infty$ this would allow a description for the action of the Steenrod Squares on any real vector bundle. Unfortunately, I am not sure there is a continous map that induces said monomorphism.

Answer 1

No, this is certainly not true. For instance, take any space $X$ for which $Sq^i:H^n(X)\to H^{n+i}(X)$ is nontrivial for some $i$ and $n$, and consider $g^*:H^*(X)\to H^*(X)$ which is $0$ in degree $n$ but the identity in degree $n+i$. Then $g^*\circ Sq^i$ is nonzero in degree $n$ but $Sq^i\circ g^*$ is $0$ in degree $n$.