fixed point projective geometry I am thinking about the following:
Let $\sigma:\mathbb C P^n\rightarrow\mathbb C P^n$ be a projectivity with $\sigma\circ\sigma=id_{\mathbb C P^n}$. I define the set of all fix points by $Fix(\sigma):=\{p\in\mathbb CP^n:\sigma(p)=p\}$.
How can I show that $Fix(\sigma)=P\cup Q$ where $P,Q$ are projetice subspaces in $\mathbb CP^n$, such that $P\cap Q=\emptyset$ and $dim(P)+dim(Q)=dim(\mathbb CP^n)-1$ ?
Can I follow that there exists an affine hyperplane $H\subseteq\mathbb CP^n$, i.e $\sigma(H)=H$, such that $\sigma_{|_A}:A\rightarrow A$ on $A:=\mathbb CP^n\setminus H$ is an affine reflection?
 A: Describing the two spaces of fixed points
The fixed points of a projective transformation correspond to the eigenspaces of its matrix. So in general you can expect $n$ distinct fixed points, but in special cases some of them might span a whole projective subspace of fixed points, and in other and even more special cases some fixed points might coincide.
Now you have the special case of an involution. Which means that the square of the diagonal matrix of eigenvalues yields some non-zero multiple the identity matrix. You can always scale the matrix of $\sigma$ in such a way that you actually get the identity matrix, and from now on I will assume that you did so. If the square of the matrix is the identity, then all eigenvalues have to be chosen from $\{-1,1\}$. So you can take $P$ to be the eigenspace corresponding to the eigenvalue $1$ and $Q$ the eigenspace for $-1$.
You get $P\cap Q=\{0\}$, i.e. the null vector, which is not a point of $\mathbb C\mathrm P^n$, butr is the only vector satisfying $1\cdot v = -1\cdot v$. You also get the fact that $\dim(P)+\dim(Q) = n+1$, as your eigenspaces span the whole $\mathbb C^{n+1}$. This is counting dimensions in the linear vector space; subtracting one to go to the projective space you get $\dim(P)+\dim(Q)=n-1=\dim(\mathbb C\mathrm P^n)-1$ as requested.
As HSN pointed out in his comment, you get into trouble if all your eigenvalues are the same, i.e. you have an identity map. So you probably should exclude the case of $\sigma=\textit{id}_{\mathbb C\mathrm P^n}$.
Because I assumed the existence of a diagonal matrix of eigenvalues for $\sigma$, you have to rule out the case that $\sigma$ cannot be diagonalized. Any non-diagonalizable matrix is still similar to a Jordan normal form, and the square of a Jordan block is never the identity matrix. So an involution will always be diagonalizable.
Interpretation as an affine reflection?

Can I follow that there exists an affine hyperplane $H\subseteq\mathbb CP^n$, i.e $\sigma(H)=H$, such that $\sigma_{|_A}:A\rightarrow A$ on $A:=\mathbb CP^n\setminus H$ is an affine reflection?

You don't know any details about the dimensions of $P$ and $Q$ separately, so you cannot be sure that one of them is a hyperplane. Only for $n=2$ you can be sure that (barring $\sigma=\textit{id}_{\mathbb C\mathrm P^2}$) you have one fixed point and one line of fixed points, so you can always break this down into an affine point reflection for which the fixed point line is treated as the line at infinity.
On the other hand, you are not actually asking about a hyperplane of fixed points, but instead about a hyperplane which is fixed as a whole. That should always be the case, I believe: you can always take one subspace as a whole and from the other a (sub-)subspace of dimension one less. The join of these two should always be a hyperplane, and as it is formed by lines which in turn are spanned by two fixed points, it should map onto itself.
I'm not so sure whether you can always interpret the remaining transformation as an affine reflection. I guess if you have a concept of affine reflection which encompasses reflections in affine spaces of arbitrary dimension, then this might work out, but for reflection in a hyperplane I believe you could construct counterexamples.
