Consider the "$m$-th power" map $f:K(\mathbb Z,n)\to K(\mathbb Z,n)$ given by $m\in \mathbb Z\cong H^n(K(\mathbb Z,n),\mathbb Z)\cong [K(\mathbb Z,n), K(\mathbb Z,n)]$. Is it true that in any degree the map $f^\*$ on integral cohomology sends any element to a multiple of $m$? It's obviously true in degree $n$ where $f^\*$ is just multiplication by $m$ but what about higher degrees?
It's enough to consider the case when $m=p$ is prime.



Action on cohomology of the power map of $K(Z,n)$
As you say, we can reduce to the case where $m=p$ is prime. By the universal coefficient theorem we know that $H^\*(K(\mathbb{Z},n);\mathbb{Z})/p$ injects in the ring $A^\*=H^\*(K(\mathbb{Z},n);\mathbb{Z}/p)$, so we just need to show that $f^\*$ acts as zero on $A^\*$ in positive degrees. The kernel of $f^\*$ is an ideal and is closed under Steenrod operations (including the Bockstein). There is a tautological class $u\in A^n$ with $f^\*(u)=pu=0$. It is known that $A^\*$ is the free unstable algebra over the Steenrod algebra generated by $u$ subject only to the relation $\beta(u)=0$. That shows that $f^\*=0$ on $A^{>0}$ as required.


