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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 $L_z= {\mathbb Z}\oplus {\mathbb Z}z\quad$. I wish to add that $P(w)=P(w,z)\quad$ is a function of two variables. Let $x(w)= \frac{P(w)-P(1/2)}{P(z/2)-P(1/2)}=x(w,z)\quad$, and $y(w)$ a suitable multiple of $P'(w)$. Then we have the Legendre form of the equation of the elliptic curve
$$y^2=x(x-1)(x-\lambda (z)),$$ where $\lambda :H \rightarrow {\mathbb P}^1\setminus \{0,1,\infty \} \quad$ is the Picard covering map. The deck transformation group is precisely $G$ (modulo $\pm 1\quad$).

Then, by the properties of the elliptic function $P(w)$, the function $P(w)-P(1/2)\quad$ has a double zero at the $2$ division point $1/2$ and hence does not vanish anywhere else. Similarly for $z/2$ and $(1+z)/2\quad$. Consequently, if we specialise $w=z/3$, then the function $x(z/3)$ does not take the value $0,1,\lambda (z) \quad$. By the lifting criterion, $$x(z/3)=x(z/3,z)= \lambda (f(z)) $$ for some $f:H \rightarrow H \quad $. Clearly, $f(z)\neq g(z)\quad$ for any $z\in H\quad$ and for any $g\in G\quad $ where $G$ is the congruence subgroup of level $2$, since their lambda values are distinct.

You can replace $z/3\quad$ by any element of the form $w=az+b \quad$ for $0< a,b < 1/2\quad $. It seems to me that not all these functions $ z\mapsto \lambda^{-1}(x(az+b))\quad $ can be fractional linear.

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 $L_z= {\mathbb Z}\oplus {\mathbb Z}z\quad$. I wish to add that $P(w)=P(w,z)\quad$ is a function of two variables. Let $x(w)= \frac{P(w)-P(1/2)}{P(z/2)-P(1/2)}=x(w,z)\quad$, and $y(w)$ a suitable multiple of $P'(w)$. Then we have the Legendre form of the equation of the elliptic curve
$$y^2=x(x-1)(x-\lambda (z)),$$ where $\lambda :H \rightarrow {\mathbb P}^1\setminus \{0,1,\infty \} \quad$ is the Picard covering map. The deck transformation group is precisely $G$ (modulo $\pm 1\quad$).

Then, by the properties of the elliptic function $P(w)$, the function $P(w)-P(1/2)\quad$ has a double zero at the $2$ division point $1/2$ and hence does not vanish anywhere else. Similarly for $z/2$ and $(1+z)/2\quad$. Consequently, if we specialise $w=z/3$, then the function $x(z/3)$ does not take the value $0,1,\lambda (z) \quad$. By the lifting criterion, $$x(z/3)=x(z/3,z)= \lambda (f(z)) $$ for some $f:H \rightarrow H \quad $. Clearly, $f(z)\neq g(z)\quad$ for any $z\in H\quad$ and for any $g\in G\quad $ where $G$ is the congruence subgroup of level $2$, since their lambda values are distinct.

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 $L_z= {\mathbb Z}\oplus {\mathbb Z}z\quad$. I wish to add that $P(w)=P(w,z)\quad$ is a function of two variables. Let $x(w)= \frac{P(w)-P(1/2)}{P(z/2)-P(1/2)}=x(w,z)\quad$, and $y(w)$ a suitable multiple of $P'(w)$. Then we have the Legendre form of the equation of the elliptic curve
$$y^2=x(x-1)(x-\lambda (z)),$$ where $\lambda :H \rightarrow {\mathbb P}^1\setminus \{0,1,\infty \} \quad$ is the Picard covering map. The deck transformation group is precisely $G$ (modulo $\pm 1\quad$).

Then, by the properties of the elliptic function $P(w)$, the function $P(w)-P(1/2)\quad$ has a double zero at the $2$ division point $1/2$ and hence does not vanish anywhere else. Similarly for $z/2$ and $(1+z)/2\quad$. Consequently, if we specialise $w=z/3$, then the function $x(z/3)$ does not take the value $0,1,\lambda (z) \quad$. By the lifting criterion, $$x(z/3)=(x(z/3,z)= \lambda (f(z)) $$ for some $f\quad $. Clearly, $f(z)\neq g(z)\quad$ for any $z\in H\quad$ and for any $g\in G\quad $ where $G$ is the congruence subgroup of level $2$, since their lambda values are distinct.

You can replace $z/3\quad$ by any element of the form $w=az+b \quad$ for $0< a,b < 1/2\quad $. It seems to me that not all these functions $ z\mapsto \lambda^{-1}(x(az+b))\quad $ can be fractional linear.

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 $L_z= {\mathbb Z}\oplus {\mathbb Z}z\quad$. I wish to add that $P(w)=P(w,z)\quad$ is a function of two variables. Let $x(w)= \frac{P(w)-P(1/2)}{P(z/2)-P(1/2)}=x(w,z)\quad$, and $y(w)$ a suitable multiple of $P'(w)$. Then we have the Legendre form of the equation of the elliptic curve
$$y^2=x(x-1)(x-\lambda (z)),$$ where $\lambda :H \rightarrow {\mathbb P}^1\setminus \{0,1,\infty \} \quad$ is the Picard covering map. The deck transformation group is precisely $G$ (modulo $\pm 1\quad$).

Then, by the properties of the elliptic function $P(w)$, the function $P(w)-P(1/2)\quad$ has a double zero at the $2$ division point $1/2$ and hence does not vanish anywhere else. Similarly for $z/2$ and $(1+z)/2\quad$. Consequently, if we specialise $w=z/3$, then the function $x(z/3)$ does not take the value $0,1,\lambda (z) \quad$. By the lifting criterion, $$x(z/3)=x(z/3,z)= \lambda (f(z)) $$ for some $f:H \rightarrow H \quad $. Clearly, $f(z)\neq g(z)\quad$ for any $z\in H\quad$ and for any $g\in G\quad $ where $G$ is the congruence subgroup of level $2$, since their lambda values are distinct.

You can replace $z/3\quad$ by any element of the form $w=az+b \quad$ for $0< a,b < 1/2\quad $. It seems to me that not all these functions $ z\mapsto \lambda^{-1}(x(az+b))\quad $ can be fractional linear.

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 $L_z= {\mathbb Z}\oplus {\mathbb Z}z\quad$. I wish to add that $P(w)=P(w,z)\quad$ is a function of two variables. Let $x(w)= \frac{P(w)-P(1/2)}{P(z/2)-P(1/2)}=x(w,z)\quad$, and $y(w)$ a suitable multiple of $P'(w)$. Then we have the Legendre form of the equation of the elliptic curve
$$y^2=x(x-1)(x-\lambda (z)),$$ 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 ellptic 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, $x(z/3)=(x(z/3,z)= \lambda (f(z)) \quad$ for some $f\quad $. Clearly, $f(z)\neq g(z)\quad$ for any $z\in H\quad$ and for any $g\in G\quad $ where $G\quad $ is the congruence subgroup of level $2$, since their lambda values are distinct.

You can replace $z/3\quad$ by any element of the form $w=az+b \quad$ for $0< a,b < 1/2\quad $. It seems to me that not all these functions $ z\mapsto \lambda^{-1}(x(az+b))\quad $ can be fractional linear.

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 $L_z= {\mathbb Z}\oplus {\mathbb Z}z\quad$. I wish to add that $P(w)=P(w,z)\quad$ is a function of two variables. Let $x(w)= \frac{P(w)-P(1/2)}{P(z/2)-P(1/2)}=x(w,z)\quad$, and $y(w)$ a suitable multiple of $P'(w)$. Then we have the Legendre form of the equation of the elliptic curve
$$y^2=x(x-1)(x-\lambda (z)),$$ where $\lambda :H \rightarrow {\mathbb P}^1\setminus \{0,1,\infty \} \quad$ is the Picard covering map. The deck transformation group is precisely $G$ (modulo $\pm 1\quad$).

Then, by the properties of the elliptic function $P(w)$, the function $P(w)-P(1/2)\quad$ has a double zero at the $2$ division point $1/2$ and hence does not vanish anywhere else. Similarly for $z/2$ and $(1+z)/2\quad$. Consequently, if we specialise $w=z/3$, then the function $x(z/3)$ does not take the value $0,1,\lambda (z) \quad$. By the lifting criterion, $$x(z/3)=(x(z/3,z)= \lambda (f(z)) $$ for some $f\quad $. Clearly, $f(z)\neq g(z)\quad$ for any $z\in H\quad$ and for any $g\in G\quad $ where $G$ is the congruence subgroup of level $2$, since their lambda values are distinct.

You can replace $z/3\quad$ by any element of the form $w=az+b \quad$ for $0< a,b < 1/2\quad $. It seems to me that not all these functions $ z\mapsto \lambda^{-1}(x(az+b))\quad $ can be fractional linear.