I am confused and curious about the meaning of the $Aut(\mathbb{CP}^n)$.

Is what is called the "linear automorphism group" of $\mathbb{CP}^n$ the same as $Aut(\mathbb{CP}^n)$? It somehow seems to me to be very non-trivial if they are the same things.

I see the statement that $Aut(\mathbb{CP}^1) = { z \mapsto \frac{az+b}{cz+d} , ad-bc \neq 0 }$ How am I supposed to interprete this statement? If $z$ is the homogeneous coordinate then its not clear to me that this map is well defined on a projective space. How does one prove this?

Is there a similar way to write down $Aut(\mathbb{CP}^2)$?

One wants to show that any two irreducible conic sections in $\mathbb{CP}^2$ are "projectively equivalent". I would like to know how this is shown. Does this mean that there exists an element of $Aut(\mathbb{CP}^2)$ that transforms one to the other? Is there a way to write down a general expression for an irreducible conic in $\mathbb{CP}^2$?

topological spaceormanifold, it has many more automorphisms. – Daniel Litt Sep 14 '11 at 21:55