Let $f \in S_2(\Gamma(N))$ be a weight 2 cusp form of full level $N > 5$. It has a Fourier expansion:
$$ f(z) = a_1 q^{1/N} + a_2q^{2/N} + \cdots. $$
Now let $S = \left(\begin{array}{cc} 0 & -1 \newline 1 & 0 \end{array}\right)$. It is not in $\Gamma(N)$, so $f|_2S \neq f$. (Here $|_2$ denotes the usual weight 2 'slash' operator.)
Nevertheless $f|_2S \in S_2(\Gamma(N))$, since $\Gamma(N)$ is normal in $SL_2(\mathbb{Z})$. Hence it too has a Fourier expansion.
What is it's Fourier expansion (in terms of the $a_i$s)?
Whatever it is, it must be such that, acting by $S$ again I get back to the $a_i$s. This suggests to me that the answer might be "complex conjugate, with possibly a minus one at the front", but I can't seem to rigorously convince myself of this (if indeed it is true), nor understand where this possible minus one comes from.
Any hints welcome.
If you conjugate $\Gamma(N)$ by $\begin{pmatrix} N & 0 \\ 0 & 1 \end{pmatrix}$ so it gets sent to a subgroup intermediate between $\Gamma_1(N^2)$ and $\Gamma_0(N^2)$ -- which corresponds to replacing $q^{1/N}$ by $q$ -- then this is just the Atkin-Lehner involution, up to some easily evaluated constant fudge factor.
This of course raises the question of explicitly computing the A-L involution on the relevant subspace of $S_2(\Gamma_1(N^2))$. I don't know a nice way of doing this, but there are certainly plenty of nasty ones.