For which $n$ is the "principal congruence subgroup" $\Gamma(n)\le \mathrm{SL}_2(\mathbb{Z})$, the subgroup consisting of matrices congruent to the identity modulo $n$, characteristic? I.e., for which $n$ is $\Gamma(n)$ stable (as a set) under all automorphisms of $\mathrm{SL}_2(\mathbb{Z})$?
No for $n$ odd, yes for $n$ even.
By a result of Hua and Reiner, the automorphism group of $SL_2(\mathbb{Z})$ is generated by (1) conjugation by $GL_2(\mathbb{Z})$ (the matrices with determinant $\pm 1$) and (2) the map $X \mapsto \epsilon(X) \cdot X$ where $\epsilon$ is the unique nontrivial character $SL_2(\mathbb{Z}) \to \pm 1$. The maps of the form (1) clearly preserve $\Gamma(n)$, so we just need to check the single map $X \mapsto \epsilon(X) \cdot X$.
We can describe $\epsilon$ as follows: $SL_2(\mathbb{Z})$ acts on the three points of $\mathbb{P}^1(\mathbb{F}_2)$, giving a map $SL_2(\mathbb{Z}) \to S_3$. Compose this with the sign character $S_3 \to \pm 1$.
If $n$ is odd, then $\left( \begin{smallmatrix} 1 & n \\ 0 & 1 \end{smallmatrix} \right) \equiv \left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right) \bmod 2$, so $\epsilon \left( \begin{smallmatrix} 1 & n \\ 0 & 1 \end{smallmatrix} \right) = 1$. So $\left( \begin{smallmatrix} 1 & n \\ 0 & 1 \end{smallmatrix} \right)$ is in $\Gamma(n)$ but $\epsilon \left( \begin{smallmatrix} 1 & n \\ 0 & 1 \end{smallmatrix} \right) \cdot \left( \begin{smallmatrix} 1 & n \\ 0 & 1 \end{smallmatrix} \right) = \left( \begin{smallmatrix} 1 & n \\ 0 & 1 \end{smallmatrix} \right)$ is not.
On the other hand, if $n$ is even then $X \equiv \mathrm{Id} \bmod n$ shows that $X$ maps to $e$ in $S_3$, so $\epsilon(X)=1$ and $\epsilon(X) X = X$ for all $X \in \Gamma(n)$.

$\begingroup$ Hmm does this imply the same result when you look at the closure of these $\Gamma(n)$'s inside $\widehat{SL_2(\mathbb{Z})}$ and continuous automorphisms? $\endgroup$ – stupid_question_bot Sep 27 '16 at 1:52

1$\begingroup$ I think you left something out. You said "let $g\in SL_2(\widehat{\mathbb{Z}})$ be $\zeta Id$ in $SL_2(\mathbb{Z}_p)$ and 1;" Did you mean "...and 1 in $\prod_{q\ne p} SL_2(\mathbb{Z}_q)$"? $\endgroup$ – stupid_question_bot Sep 27 '16 at 3:45

1$\begingroup$ Hmm, so your $X\mapsto g^{\alpha(X)}X$ only gives an automorphism of $SL_2(\widehat{\mathbb{Z}})$ right? I don't think this induces a continuous automorphism of $\widehat{SL_2(\mathbb{Z})}$ does it? For example, if you try to replace $g$ with a preimage in $\widehat{SL_2(\mathbb{Z})}$, it won't necessarily by central, right? $\endgroup$ – stupid_question_bot Sep 27 '16 at 3:52

2$\begingroup$ ${SL_2(\mathbb Z)}$ is virtually free so $\widehat{SL_2(\mathbb Z)}$ is virtually free and thus centerfree. $\endgroup$ – Ben Wieland Sep 27 '16 at 16:37

1$\begingroup$ Sorry to resurrect this post, but isn't the outer automorphism group actually $\mathbb{Z}/2\times\mathbb{Z}/2$, generated by the $\epsilon$ you describe, and by conjugation by $[[1,0],[0,1]]$ inside $\text{GL}_2(\mathbb{Z})$? Your answer is of course still valid, since $\Gamma(n)$ is normal inside $\text{GL}_2(\mathbb{Z})$ as well. $\endgroup$ – stupid_question_bot Jan 13 '17 at 3:00