Proving the Existence of an Automorphism on $\mathbb{P}^{1}$ I recently came across the following problem while reading:
Suppose that a compact Riemann surface $X$ has genus $g>1$. Let $\phi_{i}:X \rightarrow \mathbb{P}^{1}$ for $i=1,2$ be a pair of holomorphic maps, both of degree two. Prove that there is an $A \in \mathrm{Aut}(\mathbb{P}^{1})$ such that $\phi_{1}=A \circ \phi_{2}$. 
I was hoping someone could show a proof of this? I'm not really sure where to being. Thanks very much!
 A: As cant_log notes, this theorem is often stated in the form "the hyper elliptic involution is unique (when it  exists)".
The terminology arises as follows:  a complex algebraic curve (equivalently, 
compact Riemann surface) which can be written as a 2-fold branched cover of $\mathbb P^1$ is called hyperelliptic.
If $\phi: X \to \mathbb P^1$ is a double cover, then there is a uniquely determined
involution $\sigma: X \to X$ such that if $x \in X$, then $\{x ,\sigma(x)\}$
is the preimage under $\phi$ of $\phi(x)$.
The quotient $X/\sigma$ is then isomorphic to $\mathbb P^1$ (the isomorphism
being induced by $\phi$).  Thus $\sigma$ by itself determines $\phi$ up to composition
with an automorphism $A$ of $\mathbb P^1$, and so your statement can be reinterpreted as saying that when $g >1$, there is at most one involution
$\sigma$ of $X$ such $X/\sigma$ is of genus $0$.  This involution (if it exists) is
called the hyper elliptic involution of $X$.
Googling "uniqueness of hyperelliptic involution" should yield references,
I would guess.

Here is (a sketch of) one proof of uniqueness: Suppose that $\phi_1,\phi_2: X \to \mathbb P^1$
are two morphisms that don't differ just by an automorphism $A$.
Consider $\phi_1 \times \phi_2:  X \to \mathbb P^1 \times \mathbb P^1 =: Q$ (a quadric in $\mathbb P^3$).
If $\phi_1$ and $\phi_2$ don't differ by an automorphism of $\mathbb P^1$, then you can show this map is an embedding.  Its image is then a $(2,2)$-curve on $Q$.  The adjunction formula shows that an $(a,b)$-curve on $Q$ has genus $ab - a - b +1$, so in our case ($(a,b) = (2,2)$) we see that our curve $X$ has genus $1$.
