Let $G$ be an abelian group.

It seems to be a well-known fact (for example [here][1]) that $B\text{Aut}(K(G,1))$, the classifying space of the topological monoid of (unbased) self homotopy equivalences of the Eilenberg-Maclane space $K(G,1)$, has nontrivial homotopy groups $\pi_1 = \text{Aut}(G),\pi_2 = G$.

There is a fibration of topological monoids 
$$\text{Aut}_*(K(G,1)) \to \text{Aut}(K(G,1)) \to K(G,1)$$

where the first map is the inclusion of the fiber, and the second map is evaluating at the basepoint $e$ ($\text{Aut}_*$ is the topological monoid of *based* homotopy equivalences).

One can readily show that $\text{Aut}_*(K(G,1))$ is homotopy equivalent to $\text{Aut}(G)$ endowed with the discrete topology. Taking classifying spaces we get a fibration sequence

$$K(\text{Aut}(G),1) \to B\text{Aut}(K(G,1))\to K(G,2)$$

which induces the LES in homotopy 

$$\cdots \to 0 \to \pi_2(X) \to G \overset{\phi}{\to}\text{Aut}(G) \to \pi_1(X) \to 0 \to \cdots$$

where $X$ is $B\text{Aut}(K(G,1))$. 

To get the stated result, it suffices to show that the map $\phi: G \to \text{Aut}(G)$ sends $g$ to conjugation by $g$. Why is this true? I appreciate any references. 

  [1]: https://mathoverflow.net/questions/349880/classifying-space-for-fibrations-with-eilenberg-maclane-space-fibers-and-nontriv?noredirect=1&lq=1