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 the inner automorphism 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.

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.

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 the inner automorphism 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 $SL_n({\mathbb X})$. 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.

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 the inner automorphism 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.