TITLE: Example of a group which has $\text{SL}_{n}(\mathbb{Z})$ as the automorphism group
QUESTION [15 upvotes]: For the past one week, I have been trying to learn more about automorphism groups of different groups. Very recently one of my friend asked this question to me:

What is the automorphism group of $(\mathbb{Q}^{\ast},\times)$. In short, what is $\text{Aut}(\mathbb{Q}^{\ast})$?

I emailed couple of friends and got the answer as:

$\text{Aut}(\mathbb{Q}^{\ast})$ is isomorphic to the automorphism group of a free abelian group of countable rank. In particular, it will contain $\text{GL}(n,\mathbb{Z})$ for all $n$.

My question would be :


Can we realize $\text{SL}_{n}(\mathbb{Z})$ to be the automorphism group of some group?


Are there groups which are which are "very difficult" to be realized as the automorphism group of a certain group.


So suppose someone comes and asks me: Is $S_{3}$ or $\text{GL}_{2}(\mathbb{Z})$ the automorphism group of some group, then how can I answer the question? I am particularly interested in seeing how to think for a solution.

REPLY [6 votes]: Another not quite answer: In the beautiful paper
Automorphism groups of polycyclic by finite groups and arithmetic groups, by Oliver Baues and Fritz Grunewald, 
Publ. Math. IHES 104 (2006), no. 1 (arXiv version, MathReviews, journal version)
the authors show that in any cases the automorphism groups of polycyclic-by-finite groups are "arithmetic" (basically subgroups of $SL(n, \mathbb{Z})$ (results of this sort have actually been know for quite a while for narrower classes of groups -- check the references of the paper).