Isometries of complex space

I am trying to find the isometries of the complex space $\mathbb{P}^n\mathbb{C}$ which is given a riemannian metric using the Hopf fibration. What I proved so far is that every matrix $A \in U(n+1)$ induces an isometry of the complex space which seems to be orientation preserving. Moreover using the transitivity of the action of the unitary group on the sphere $2n+1$ dimensional I proved that (if the claim that the above isometry is orientation preserving is true) the group of orientation preserving isometries is given by:

$$U(n+1) / U(1) = SU(1)$$

What I also proved is that an isometry of the $2n+1$ sphere descend to an isometry of the complex space if and only if the corresponding matrix lies in $SO(2n+2)$. Now the questions are:

1. Are all the isometries of the complex plane orientation preserving?
2. Do all the isometries of the complex space arise as maps induced by isometries of the $2n+1$ dimensional sphere?

Thank you.

EDIT: For me an isometry is a map $f: \mathbb{P}^n\mathbb{C} \rightarrow \mathbb{P}^n\mathbb{C}$ such that for every $P \in \mathbb{P}^n\mathbb{C}$ the differential $d_Pf$ is an isometry of $T_P \mathbb{P}^n\mathbb{C}$ in $T_{f(P)} \mathbb{P}^n\mathbb{C}$ with the metric specified at the beginning of the post.

• I edited the post, thank you. – Federico Dec 28 '17 at 19:51
• Okay. When you talk about "differential" (which is a linear map between tangent spaces), I gather the map is $\mathbb{C}$-linear? Every invertible $\mathbb{C}$-linear map is orientation preserving. Consequently, every biholomorphic map between complex manifolds is orientation preserving. – user357151 Dec 28 '17 at 19:58
• No, when looking at $\mathbb{P}^n\mathbb{C}$ as a riemannian manifold I am using its structure as a real manifold. – Federico Dec 28 '17 at 20:03