Theorem: The infinite cyclic group, written $C_{\infty}$, cannot be the automorphism group of any group.
Also known as $\mathbb{Z}$ (under addition), this is the canonical example of such a group (and it seems my proof works for all cyclic groups of odd order). The proof is easy, and consists of two lemmata. I shall leave you to pin the lemmata together and get the result (Hint: what are the inner automorphisms isomorphic to?).
Lemma 1: If $G/Z(G)$ is cyclic then $G$ is abelian.
Proof: This is a standard undergrad question, so I'll let you figure out the proof for yourself.
Lemma 2: If $G\not\cong C_2$ is abelian then $\operatorname{Aut}(G)$ has an element of order two.
(Here, $C_2$ is the cyclic group of order two. Note that this group has trivial automorphism group.)
Proof: The negation map $n: a\mapsto a^{-1}$ is non-trivial of order two unless $G$ comprises of elements of order two. If $G$ consists only of elements of order two then $G$ is the direct sum of cyclic groups of order two, $$G\cong C_2\times C_2\times\ldots$$ See this question for why. Finally, because $G\not\cong C_2$ there are at least two copies of $C_2$, and so we can switch them (and "switching" has order two). Note: This automorphism depends on the Axiom of Choice (this was pointed out by Arturo Magidin in an answer to a question on the group pub forum mailing list).