Note that $\Omega$ of such a map would be a map $K(\mathbb{Z},3)\rightarrow S^3$, which I think is nullhomotopic, so if there is a nontrivial map like you want it'd act trivially on $\pi_4$, $H^4$ and so all cohomology.

I wouldn't say "very hard to describe", there's a nice PL map, triangulating both squares into six triangles ;)

@Gerhard the estimate is lower, because here we count an isomorphism class weighted by $1/|G|$ instead of $1$, where $G$ is its automorphism group. Think about the case where we want to count all graphs with two vertices and one edge: the estimate suggests $1/2!$, the correct answer is $1$.

It is $1+kw_1+\frac{k(k-1)}{2}w_1^2$, which is $1$ for $k=4n$ or $k=4n+1$.

(Say action on $H_n(\bullet; \mathbb{Z})$ to get around basepoints, although for $n\geq 2$ that's not an issue)

Associating to a map its action on $\pi_n$, you get a map of sets $Aut(K(A,n)) \rightarrow Aut(A)$, and since two homotopic maps of induce the same map on $\pi_n$, it is constant on connected components of $Aut(K(A,n))$, therefore continuous when the right side is endowed with the discrete topology.

David may have confused you by writing the normal subgroup of the semidirect product on the right side in his question.

We do have a group homomorphism in the other direction, from $Aut(K(A, n)) \rightarrow Aut(A) $, which splits. By the long exact sequence of homotopy groups, the fiber is a $K(A, n)$.

Alexander-Poincare duality (as in Hatcher, Algebraic Topology, Thm. 3.44) shows that for any embedding of $Gr(k,n)$ into $Gr(k,n+1)$, the cohomology of the complement is related by a long exact sequence to the homology of $Gr(k,n)$ and the cohomology of $Gr(k,n+1)$. This suggests that the topology of the complement is very tightly constrained. To prove that there's no other possible complement is probably very hard, though. Still, this excludes the point and many other spaces essentially simpler than $Gr(k-1,n)$.

Nope, not true in general: Take a countable union of copies of $S^1$ and define a map $\cup S^1\rightarrow S^1$ by the double cover $S^1\rightarrow S^1$ on each summand. The fiber is countably many points, the total space is a product of countably many points with $S^1$, but there's no section.

To add a third description (which is the one I use in my comment): If you embed $\mathbb{Z}/p^n\rightarrow \mathbb{Z}/p^{n+1}$, sending $1$ to $p$, you basically add in $1/p$ of the generator. $\mathbb{Z}/p^\infty$ is defined as the colimit over this sequence of maps, so it is filtered by cyclic subgroups of order $p^n$, each of which is $p$ times the next. That connects directly to both of Sean's perspectives. If you're familiar with how towers in the Adams spectral sequence encode cyclic groups of order $p^n$, the right intuition here is a tower that extends infinitely far down.

In general, a map that induces isos on homology with field cofficients induces isos on integral homology. A quick rundown: For $A\rightarrow B\rightarrow C$ a short exact sequence of abelian groups, if it works for two of these coefficients, it works for the third - that's the five-lemma. Start from $\mathbb{F}_p$ to inductively prove it for all $\mathbb{Z}/p^n$. Then it also follows for their colimit, $\mathbb{Z}/p^{\infty}$. But $\mathbb{Q}/\mathbb{Z}$ decomposes into copies of $\mathbb{Z}/p^{\infty}$, so we're done by $\mathbb{Z}\rightarrow \mathbb{Q}\rightarrow \mathbb{Q}/\mathbb{Z}$.

The $Tor_1$ isn't zero: There's an exact sequence $Tor_1(\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) \rightarrow Tor_1(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \rightarrow \mathbb{Z}\otimes \mathbb{Q}/\mathbb{Z} \rightarrow \mathbb{Q}\otimes \mathbb{Q}/\mathbb{Z}$. The right- and leftmost terms are 0, so the Tor is just $\mathbb{Q}/\mathbb{Z}$.