Is every group the automorphism group of a group?

Suppose $G$ is a group. Does there always exist a group $H$, such that $\operatorname{Aut}(H)=G$, i. e. such that $G$ is the automorphism group of $H$?

EDIT: It has been pointed out that the answer to the above question is no. But I would be much more pleased if someone could give an example of such a group $G$ not arising as an automorphism group together with a comparatively easy proof of this fact.

Theorem. The following cyclic groups cannot be the automorphism group of any group:

1. The infinite cyclic group $C_{\infty}$ (also known as $\mathbb{Z}$), and
2. Cyclic groups $C_{n}$ of odd order (also known as $\mathbb{Z}_n$ or $\mathbb{Z}/n\mathbb{Z}$).

The proof is relatively straight forward, with a subtlety at the end, 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, applying the Axiom of Choice, $G$ 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).

The subtlety I mentioned at the start is the use of Choice in the proof of Lemma 2. If we do not assume Choice that it is consistent that there exists a group $G$ of order greater than two such that $\operatorname{Aut(G)}$ is trivial. This was (first) proven by Asaf Karagila in an answer to this MSE question.

Evidently this is false even if $H$ is required to be finite.

I think the argument would be very difficult if $G$ was allowed to be infinite, especially if $G$ was not finitely generated.

EDIT: Here is a cool somewhat related result.