No for $n$ odd, yes for $n$ even.
By a result of Hua and Reiner, the outer automorphism group of $SL_2(\mathbb{Z})$ is $\mathbb{Z}/2$, so we just need to check whethergenerated by $\Gamma(n)$ is preserved(1) conjugation by one representative. The representative Hua$GL_2(\mathbb{Z})$ (the matrices with determinant $\pm 1$) and Reiner give is(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 $X \in \Gamma(n)$.