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.