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Let $\Gamma=\Gamma_o(N)$ be the congruence subgroup. Let $f\in C^\infty(\Gamma\backslash GL(2,R)/SO(2,R)R^*)$ be a Maass form. How shall we define its dual(contragredient) Maass form $f'$?

If $\Gamma=SL(2,Z)$, it is known that the dual Maass form should be $f'(z)=f(\omega (z^t)^{-1}\omega^{-1})$, where $$\omega=\begin{pmatrix}0&-1\\\ 1&0\end{pmatrix}.$$

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    – GH from MO
    Commented Aug 19, 2018 at 22:59

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If $f(z)$ is a Maass form of weight zero and level $N$ on the upper half-plane, then the dual form is $f\left(\frac{-1}{Nz}\right)$. One can transfer this statement to $GL(2,R)$ using Iwasawa coordinates.

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