This is an answer to part of question 1. Such a function $f:H \rightarrow H$ does exist. To see this, let $z\in H$ and $P(w)$ be the meromorphic function on the plane which is the Weierstrass $P$ function corresponding to the lattice $L_z= {\mathbb Z}\oplus {\mathbb Z}z$. Let $x(w)= \frac{P(w)-P(1/2)}{P(z/2)-P(1/2)}$, and $y(w)$ a suitable multiple of $P'(w)$. Then we have the Legendre form of the elliptic equation $$y^2=(x(x-1)(x-\lambda (z)),$$ where $\lambda :H \rightarrow {\mathbb P}^1\setminus \{0,1,\infty \}$ is the Picard covering map. The deck transformation group is precisely $G$ (modulo $\pm 1$). Then, by the properties of the elellpitic function $P(w)$, the function $P(w)-P(1/2)$ has a double zero at the $2$ division point $1/2$ and hence does not vanish anywhere else. Similarly for $z/2$ and $(1+z)/2$. Consequently, if we specialise $w=z/3$, then the function $x(z/3)$ does not take the value $0,1,\lambda (z)$. By the lifting criterion, $x(z/3)=\lambda (f(z)$ for some $f$. Clearly, $f(z)\neq g(z)$ for any $z\in H$ and for any $g\in G$ where $G$ is the congruence subgroup of level $2$, since their lambda values are distinct.
You can replace $z/3$ by any element of the form $w=az+b$ for $0< a,b < 1/2$. It seems to me that not all these functions $ z\mapsto x(az+b)$ can be fractional linear.