Let $\Gamma$ be a finite index subgroup of $\Gamma(1)=SL_2(\mathbf{Z})$ and $f$ a modular function for $\Gamma$. By this I mean a meromorphic function defined on the upper half-plane $f: \mathfrak{h} \to \mathbf{C}$, satisfying a growth condition at infinity, which is invariant by $\Gamma$: for every $\gamma \in \Gamma$, $f(\gamma(\tau))=f(\tau)$.
My question is: what can be said about the action of the full modular group $\Gamma(1)$ on $f$? For $\gamma \in \Gamma(1)$, can we always express the function $\tau \mapsto f(\gamma(\tau))$ in terms of $f$?
For example, take the modular lambda function $\lambda$. It is an everywhere holomorphic modular function (or better, modular form of weight zero) for $\Gamma(2) = \{\gamma \in \Gamma(1) \mid \gamma \equiv id \mod 2\}$. It can be proven that \begin{align*} \lambda(-1/\tau) = 1-\lambda(\tau) \end{align*} and \begin{align*} \lambda(\tau +1 ) = \frac{\lambda(\tau)}{1-\lambda(\tau)} \end{align*} so that we actually know how $\lambda$ transforms by $\Gamma(1)$, and not just $\Gamma(2)$. I'm asking if this is general.
No, this cannot always be done. For instance, the function $$f(z) = \left(\Delta(5z) / \Delta(z)\right)^{1/4}$$ is a modular function of level $\Gamma_0(5)$, but if you act on it by an element of $\Gamma(1)$ that doesn't normalise $\Gamma_0(5)$, such as $z \mapsto -1/z$, the result won't be $\Gamma_0(5)$-invariant, so it certainly can't be expressed as a rational function in $f$.