Let $$\pi:S^{2n+1}\to S^{2n+1}/S^1=\mathbb{P}^n\mathbb{C}$$ be the Hopf fibration. I already know that $d_z\pi$ is a vector space isomorphism for every $z\in S^{2n+1}$ (when restricted to the orthogonal space to the orbit). We'd like to define a riemannian structure on $\mathbb{P}^n\mathbb{C}$ by push-forwarding the riemannian structure of $S^{2n+1}$ on each tangent space through $d_z \pi$. Basically we want to define the Fubini metric as:
$$\mathsf{fub}_{[z]}(u,v):=\mathsf{sph}_z((d_z\pi)^{-1}(u),(d_z\pi)^{-1}(v)),$$
where $\mathsf{sph}$ is the standard spheric metric. The only problem with this definition is that it may depend on the choice of the representative $z$ of the equivalence class $[z]$. So let $e^{i\theta}z$ be a new representative:
$$\mathsf{sph}_{e^{i\theta}z}((d_{e^{i\theta}z}\pi)^{-1}(u),(d_{e^{i\theta}z}\pi)^{-1}(v))=\langle (d_{e^{i\theta}z}\pi)^{-1}(u), d_{e^{i\theta}z}\pi)^{-1}(v) \rangle$$
where $\langle -,- \rangle$ is the standard euclidean inner product on $\mathbb{C}^{n}\cong \mathbb{R}^{2n}$ (the fixed isomorphism is $(x_1+iy_1,...,x_n+iy_n)\to (x_1,...,x_n,y_1,...,y_n)$). My problems would be over if I had an euclidean isometry $I:\mathbb{C}^n\to \mathbb{C}^n$ such that:
$$(d_{e^{i\theta}z}\pi)^{-1}=I\circ (d_z\pi)^{-1}$$
so basically I'd like to prove that:
$$(d_{e^{i\theta}z}\pi)^{-1}\circ (d_z\pi)$$
is an euclidean isometry. Is there an elegant way of doing this without dwelving too much into coordinates? Or in alternative, is there a better approach?
Let $\iota\colon S^{2n+1} \to \Bbb C^{n+1}$ be the inclusion map. Let $f_{\lambda}\colon \Bbb C^{n+1} \to \Bbb C^{n+1}$ and $\bar{f}_{\lambda}\colon S^{2n+1}\to S^{2n+1}$ both be the multiplication by $\lambda \in \Bbb S^1$ (but not on the same space). Both are diffeomorphisms. One readily checks that the following commutative diagram holds $$\require{AMScd} \begin{CD} S^{2n+1} @>{\iota}>> \Bbb C^{n+1}\\ @V{\bar{f}_{\lambda}}VV @VV{f_{\lambda}}V \\ S^{2n+1} @>{\iota}>> \Bbb C^{n+1} \end{CD}$$ that is, that $\iota\circ \bar{f}_{\lambda} = f_{\lambda}\circ \iota$.
Let $g$ be the natural metric on $\Bbb C^{n+1}$. It is clear that $f_{\lambda}$ is an isometry, so that $f_{\lambda}^*g = g$. Let $h=\iota^*g$ be the induced metric on the sphere. Then $$ \bar{f}_{\lambda}^* h = \bar{f}_{\lambda}^*\iota^*g = (\iota\circ \bar{f}_{\lambda})^*g= (f_{\lambda}\circ \iota)^*g = \iota^* f_{\lambda}^*g = \iota^*g = h, $$ that is, $\bar{f}_{\lambda}$ is an isometry for $h$.