Suppose we pick a natural isomorphism between $H^n(-;G)$ and $\langle -, K(G,n)\rangle$, when does an element of $H^n(K(G,n),G)=\operatorname{Hom}(G,G)$ correspond to a self map of $K(G,n)$ that has a homotopy inverse?
I believe I can answer for the case $G = \mathbb{Z}$ and the "standard" natural isomorphism which is given by $[f] \in \langle X, K(G,n)\rangle \rightarrow f^*([\sigma])$ where $\sigma$ is a cocycle whose value on cycles is given by using the Hurewicz isomorphism to interpret the cycles as elements of $G$.
In this case, I believe $1,-1 \in \mathbb{Z}=\operatorname{Hom}(\mathbb{Z},\mathbb{Z})$ are the only elements of $H^n(K(G,n),\mathbb{Z})$ that correspond to invertible self maps. These correspond to invertible self maps because either $1$ or $-1$ will be the image of the identity under the natural isomorphism (depending on how you identify $\pi_n(K(\mathbb{Z},n))$ and $\mathbb{Z}$), and the other will correspond to the map $x \rightarrow -x$ where the negation is that of loops. This latter one is an involution.
To show that these are the only invertible (up to homotopy) self maps, I believe we can adopt a right unital monoidal action of $\operatorname{Aut}(K(G,n))=\langle K(G,n),K(G,n)\rangle$ on itself, where as a monoid the multiplication is composition and as an abelian group the operation is pointwise composition of loops.
Distributivity from the right holds: ([f]+[g])[h]=[f][h]+[g][h]. This is standard (it is needed to prove the maps into Eilenberg-MacLane spaces even form a functor). Obviously, the action is unital.
Now multiplication on the right by an element of $\operatorname{Aut}(K(G,n))$ is a group endomorphism, and if the element is invertible the homomorphism must have a left inverse. When $G=\mathbb{Z}$ this implies it is an automorphism.
The only automorphisms of $\mathbb{Z}$ are multiplication by $1,-1$. These correspond only to the maps $[x \rightarrow x],[x \rightarrow -x]$ because we can look at the action of our map on the identity and conclude the homotopy type of the map.
Is it possible to adapt this strategy to work for things like finite groups? Free abelian groups? If an element of $\operatorname{Hom}(G,G)$ represents an invertible self map for some $n$ must it for all $n$? Or can anyone just answer the question in the title?
This is much easier than you're making it. The standard isomorphism $H^n(K(G,n),G)\to\operatorname{Hom}(G,G)$ can be described as follows: for any $(n-1)$-connected space $X$, we have natural isomorphisms $$H^n(X,G)\cong \operatorname{Hom}(H_n(X,\mathbb{Z}),G)\cong \operatorname{Hom}(\pi_n(X),G)$$ where the first isomorphism is by the universal coefficient theorem and the second isomorphism is by Hurewicz. We now use this naturality for a map $f:K(G,n)\to K(G,n)$ to get a commutative diagram $$\require{AMScd} \begin{CD} H^n(K(G,n),G) @>{\cong}>> \operatorname{Hom}(\pi_n(K(G,n)),G)\\ @V{}VV @V{}VV \\ H^n(K(G,n),G) @>{\cong}>> \operatorname{Hom}(\pi_n(K(G,n)),G) \end{CD}$$ where the horizontal maps are the standard isomorphism and the vertical maps are induced by $f$. Now start with the canonical class in the top left. Going down, this by definition maps to the class in $H^n(K(G,n),G)$ corresponding to $f$, and so then going right we get the corresponding element of $\operatorname{Hom}(G,G)$. But if we instead go right first, we get the identity element of $\operatorname{Hom}(G,G)$ in the top right, which then maps down to the element of $\operatorname{Hom}(G,G)$ induced by $f$ on $\pi_n$.
In other words, the element of $\operatorname{Hom}(G,G)$ corresponding to $f$ is just the induced map on $\pi_n$. By the Whitehead theorem, $f$ is a homotopy equivalence iff it induces an isomorphism on $\pi_n$, i.e. iff the corresponding element of $\operatorname{Hom}(G,G)$ is invertible.