Thanks for your answer Anton. I don't understand your argument however. Do you mind being a bit pedantic for me and spelling it out? So, from what you wrote, the space of self-maps of $K(A,n)$ in pointed spaces is equivalent to the discrete set $End(A).$ How do I go from here to concluding that we have the semi-direct product decomposition I'm after of the entire automorphism space?

I see how both parts act. My question is how do we see this is everything?

The particular case you care about is covered by the fact that $Shv(C/c)\simeq Shv(C)/y(c),$ where $y$ is the Yoneda embedding. With this insight, the geometric morphism you seek IS in HTT; it's an etale geometric morphism. The fact I claimed is Proposition 2.2.1 here: arxiv.org/abs/1312.2204

I removed my answer since I misread something in Higher Algebra.

Thanks for the comments. I'm having trouble tracking down exactly which paper this occurs in. Any idea the title? Thanks again!

@Sam: Thanks for pointing this out. I had even glanced at that paper earlier this year but had forgotten. It's reasonable to expect from this that non-seperated schemes over $\mathbb{C}$, even if of finite type, my fail to have the homotopy type of a finite CW-complex. However, the example you give turns out to be a $K(\mathbb{Z},1),$ so the question still remains if 2.) holds...