Automorphisms of $SL_n(\mathbb{Z})$ What is the group of outer automorphisms of $SL_n(\mathbb{Z})$.  I wanted to understand semidirect products of the form $SL_n(\mathbb{Z})\rtimes_\varphi \mathbb{Z}$ and its isomorphism type depends only on $[\varphi]\in Out(SL_n(\mathbb{Z})$. There is always the conjugate inverse, which is clearly not an inner automorphism, as it doesn't preserve the minimal polynomial of the matrix (at least for $n\ge 3$). Are there any other outer automorphisms ?
 A: As I suggested in my short comment, this kind of question has been around for a long time and has led to a vast amount of literature.   It probably starts with work over fields by Schreier and van der Waerden in the 1920s, then considerable work by Dieudonne, O'Meara, and many others.    Two indications about what's out there are a survey by O'Meara The integral classical groups and their automorphisms (1969) and a short paper by me dealing from the algebraic group viewpoint with groups like $SL_n(\mathbb{Q})$,  On the automorphisms of infinite Chevalley groups, Canad. J. Math. 21 1969 908–911, the latter probably available online by now.  
These are listed on MathSciNet, but I couldn't display the links there for some reason.  I'll try to suggest more focused literature on your question when I get time today.
P.S.  Ed Formanek has pointed to a basic early paper by Hua and Reiner here.  Like many other papers on related automorphism groups, the emphasis is on identification of special types of automorphisms which suffice to generate the whole group: inner, "field" or "ring" (as in complex conjugation and the like), "graph" (as in transpose-inverse map for special linear groups), "diagonal" (as in the use of conjugation by diagonal matrices not of determinant 1 to produce automorphisms of a subgroup).   Naturally the ring or field automorphisms play no role over the rational integers or rational numbers.   
A: Conjugating by an element $Q \in GL_n(\mathbf{Z})$ is usually not inner, furthermore the graph automorphism $A \mapsto (A^T)^{-1} $ is not inner.
The phrase to google for is "automorphisms of Chevalley groups over rings" which in particular turns up recent work of Bunina.
By invoking Margulis superrigidity, for $n \geq 3$, any automorphism of $SL_n(\mathbf Z)$ extends to a rational automorphism of $SL_n(\mathbf Q)$. To show that the above-mentioned examples then are the only ones, one calculates the stabilizer of $SL_n(\mathbf Z)$ in $GL_n(\mathbf Q)$. 
A: One possible approach (perhaps not the most elementary one) is to use Margulis' Superrigidity theorem. $\Gamma=SL_n({\mathbb Z})$ is a lattice in the simple Lie group $SL_n({\mathbb R})$ which has rank $n-1 \ge 2$ when $n \ge 3$. Now, if $\phi: \Gamma \to SL_n({\mathbb R}$ is a homomorphism such that $\phi(\Gamma)$ is Zariski dense, the above-mentioned theorem guarantees that $\phi$ extends to a {\it rational} homomorphism $\phi': SL_n({\mathbb R}) \to SL_n({\mathbb R})$. This condition is automatically satisfied here, so the problem boils down to characterizing the rational automorphism of $SL_n({\mathbb R})$. This is easy, because you can show that conjugations with matrices in $GL_n(\mathbb R)$ form a subgroup of index 2 there, and generate the full group with with $x \mapsto ^tx^{-1}$. You have to check which one of these will stabilize 
$SL_n({\mathbb Z})$. It is certainly in $SL_n({\mathbb Q})$ which is the commensurability subgroup. I have not checked this, but I guess that the normalizer here is exactly $SL_n({\mathbb Z})$.
The case of $SL_2(\mathbb Z)$ is probably more complicated. Superrigidity is certainly false, at least for some subgroups of finite index in $SL_2(\mathbb Z)$, since $SL_2(\mathbb Z)$ has free subgroups of finite index with a lot of automorphisms. If they do extend to $SL_2$ (I am not sure if this is the case) they you get an abundance of automorphisms that are not algebraic, and perhaps you cannot easily classify them. 
A: See L. K. Hua and I. Reiner, "Automorphisms of the unimodular group", Trans. Amer. Math. Soc.
71 (1951), 331-348.
