Isomorphisms of $\mathbb P^1$ 
Prove that every isomorphism of $\mathbb P^1$ (over an algebrically closed field $\mathbb K$) is of the form 
  $$
\phi(x_0: x_1) = (ax_0+bx_1 : cx_0 + dx_1)
$$
  where $\begin{pmatrix} a & b \\c & d\end{pmatrix} \in GL(2, \mathbb K).$

There are some hints; I show you what I've done.  
1. Show that the action of $PGL(2, \mathbb K)$ on $\mathbb P^1$ is transitive. 
Well, I think this is quite intuitive, but I can't find a rigorous proof. Let us pick two distinct points $x=(x_0:x_1)$ and $y=(y_0 : y_1)$. I want to find an invertible matrix $A$ s.t. $Ax = y$. It is easy to see that this system (the unknowns are the entries of $A$) has always solutions (and maybe one can find one solution s.t. $ad-bc \ne 0$). My question is: is there an easier way to see that this action is transitive? Is it enough to say "It's trivially true in the affine case so it must be true in the projective space as well"? 
2. Using the fact that every isomorphism of $\mathbb A^1$ is of the form $t \mapsto at+b$ (for some $a \ne 0$), prove the claim for morphisms s.t. $\phi(0:1)=(0:1)$.
Suppose $\phi$ is a morphism s.t. $\phi(0:1)=(0:1)$. If we restrict $\phi\vert_{U_0}$ we obtain a morphism of $U_0 \simeq \mathbb A^1$, hence it must of the form 
$$
\phi(1:x) = (1:ax+b) 
$$
i.e. 
$$
\phi(x_0:x_1) = (x_0:ax_0+bx_1) 
$$
Is this correct? Why shall I assume that $(0:1) \mapsto (0:1)$?
3. Conclude. 
I don't see how to conclude at this point, I'm puzzled. Could you please help me? 
Thanks. 
 A: *

*It is enough to show that  $(0,1)$ can be moved to any point. This is because if I want to move $x$ to $y$, I just need to move $x$ to $(0,1)$ by the inverse that moves $(0,1)$ to $x$, and then move $x$ by the map that moves $(0,1)$ to $y$. But this is clear because any $y$ can be represented by say $(a,b)$ and then the matrix that moves $(0,1)$ to $(a,b)$ is given by
$$\left(\begin{array}{cc} c_1 & a \\ c_2 & b \end{array}\right)$$
where $c_1,c_2 \in k$ chosen so that $c_1b - ac_2 \neq 0$. Now why do such $c_1,c_2$ have to exist?
As for the last part of your question, I don't understand what you mean when you say "It's true in the affine case thus true in the projective case". What is true in the affine case? I would be happy to edit my answer.

*For your second question, I believe your answer is correct. The key point as you realize is that if $\phi$ fixes the point at infinity then it restricts to an open subset isomorphic to $\Bbb{A}^1$.

*The reason why we first prove the case $\phi(0,1) = (0,1)$ is this. Suppose in the general case that you are given any isomorphism $\varphi$ of $\Bbb{P}^1$; i.e. suppose $\varphi(0,1) = (c,d)$. By the first part you know there is an element of $\text{PGL}(2)$ that moves $(c,d)$ to $(0,1)$ and say it is given by some $\rho$. The composite $\rho \circ \varphi$ is then still an automorphism of $\Bbb{P}^1$ (because $\rho$ is) and so immediately we know 
$$\rho \circ \varphi = \phi$$ 
for some $\phi$ that is of the form $\phi(x_0,x_1) = (x_0,ax_0 + bx_1)$ by (2). Then from here we get $\varphi = \rho^{-1} \circ \phi$ and so..... (conclude since this is a homework problem).
