A friend recently told me the following two facts, for which he cannot recall a proof or a reference
(but he remembers seeing them in the literature):


1. Let $f$ be a holomorphic function mapping the upper half-plane $H$ into itself. Let $G$ be the
group of fractional linear transformations $(az+b)/(cz+d)$ where $ad-bc=1$, $a,d$ are odd integers
and $b,c$ are even integers. Suppose that for every $g\in G$ and for every $z\in H$,
$f(z)$ is not equal to $g(z)$. Then $f$ is fractional-linear. (Or maybe such $f$ just does not exist).
2. Let $f$ be the same as before. Suppose that for all integers $m,n$ and all $z\in H$
we have $f(z)\neq mz+n$. Then $f$ is fractional-linear.


I will appreciate any relevant reference or any other information. 


EDIT: There is no $f\in Aut(H)$ that satisfies the condition of Problem 1. This implies that $f$ constructed by
 Aakumadula is NOT fractional-linear.


To prove this, we write $(az+b)/(cz+d)=(xz+y)/(uz+t)$, where $a,b,c,d$ are given real numbers,
and we want to find integers $x,y,u,t$, where $x,t$ are odd, and $y,u$ are even, so that this has non-real roots
$z$. This is to show that certain quadratic form in $a,b,c,d$ is indefinite. And this is performed
by an elementary calculation.



Analytic function avoiding elements of the modular group
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. 


