Let $R$ be the matrix $\left[\begin{smallmatrix}-1 & 0\\ 0 & 1\end{smallmatrix}\right]$. Let $s$ be the involution of $SL_2(\mathbb{Z})$ given by $s(A) = RAR$
Let $\mathbb{H}$ be the upper half plane, and $\mathbb{H}^-$ the lower half plane. We have an action of $SL_2(\mathbb{Z})$ on $\mathbb{H}\sqcup \mathbb{H}^-$ by $\gamma\tau = \frac{a\tau+b}{c\tau+d}$, if $\gamma = \left[\begin{smallmatrix}a & b \\c & d \end{smallmatrix}\right]$.
Let $\Gamma\le SL_2(\mathbb{Z})$ be a finite index subgroup. We have the Riemann surface $\mathbb{H}/\Gamma$, as well as its "mirror image" $\mathbb{H}/s(\Gamma)$.
In general, these two curves are antiholomorphically homeomorphic to each other: The antiholomorphic automorphism of $\mathbb{H}$ given by $\tau\mapsto R\overline{\tau} = -\overline{\tau}$ induces an antiholomorphic homeomorphism $$\mathbb{H}/\Gamma\rightarrow\mathbb{H}/s(\Gamma)$$
In certain cases, e.g. if $s(\Gamma) = \Gamma$ (equivalently, $R\Gamma R = \Gamma$), we even have an isomorphism (i.e., holomorphic homeomorphism) $\mathbb{H}/\Gamma\cong\mathbb{H}/s(\Gamma)$. This is the case for example for the common congruence subgroups $\Gamma(n),\Gamma_1(n),\Gamma_0(n)$.
If $s(\Gamma)\ne\Gamma$, could there still be an isomorphism $\mathbb{H}/\Gamma\cong\mathbb{H}/s(\Gamma)$?
Bonus question: Both curves $\mathbb{H}/\Gamma$ and $\mathbb{H}/s(\Gamma)$ map to $\mathbb{H}/SL_2(\mathbb{Z})$, and hence the standard $\mathbb{Q}$-structure on $\mathbb{H}/SL_2(\mathbb{Z})$ gives $\mathbb{H}/\Gamma$ and $\mathbb{H}/s(\Gamma)$ the structure of algebraic curves over some number field. Does the fact that they are mirror images (anti-holomorphically homeomorphic) say anything about their arithmetic structures? (maybe some kind of twist by complex conjugation?)