Different(?) measures on sphere Let us consider the real sphere $S^{2n-1}:=\{x=(x_1,\ldots,x_{2n})\in\Bbb R^{2n}\mid\lVert x\rVert=1\}$ and the complex sphere $S_\Bbb C^{n-1}:=\{z=(z_1,\ldots,z_n)\in\Bbb C^n\mid\lVert z\rVert=1\}$. Then we have an isomorphism 
\begin{align*}
\Phi:S_\Bbb C^{n-1}\rightarrow S^{2n-1},\quad z\mapsto x
\end{align*}
where $z_1=x_1+ix_{n+1},z_2=x_2+ix_{n+2},\ldots,z_n=x_n+ix_{2n}$. Now consider the inner product 
\begin{align*}
\langle f,g\rangle_{L^2(S_\Bbb C^{n-1})}:=\int_{S_\Bbb C^{n-1}}f(z)\overline{g(z)}dz
\end{align*} 
of functions $f,g$ on $S_\Bbb C^{n-1}$ where $dz$ is the $U(n)$-invariant measure on $S_\Bbb C^{n-1}$ and let 
\begin{align*}
\langle f,g\rangle_{L^2(S^{2n-1})}:=\int_{S^{2n-1}}f(x)\overline{g(x)}dx
\end{align*} 
for two functions $f,g$ on $S^{2n-1}$ where $dx$ is the $O(2n)$-invariant measure on $S^{2n-1}$. My question is: Is it true that, after normalizing the measures, it holds for functions $f,g$ on $S^{2n-1}$ that
\begin{align*}
\langle f,g\rangle_{L^2(S^{2n-1})}=\langle f\circ\Phi, g\circ\Phi\rangle_{L^2(S_\Bbb C^{n-1})}?
\end{align*}
If not, is there any correlation like $\langle f,g\rangle_{L^2(S^{2n-1})}=0\Rightarrow \langle f\circ\Phi, g\circ\Phi\rangle_{L^2(S_\Bbb C^{n-1})}=0$? Thanks for any help in advance.
 A: Yes, that holds. The normalized $(2n-1)$-dimensional Lebesgue measure is also the same.
For even $n$ it's also the unique normalized invariant measure for the quaternion unit group $Sp(n/2).$
I think from the wording of your question that you are already aware of the following result for homogeneous spaces (or perhaps some more general result).

Given a compact Lie group $G$ acting transitively on a compact manifold $M,$ there exists a unique $G$-invariant probability measure on $M.$

I find using an explicit $\Phi$ obscures what is going on. Sorry if this is confusing, but I will take $\mathbb R^{2n}$ to have the same underlying real vector space and norm as $\mathbb C^n,$ so $\Phi$ is the identity. (The inner products are slightly different: $\langle x,y\rangle_{\mathbb R^{2n}}=\mathrm{Re}\langle x,y\rangle_{\mathbb C^{n}}.$)
Every unitary transformation of $\mathbb C^{n}$ is an orthogonal transformation of $\mathbb R^{2n}$: a unitary transformation is a $\mathbb C$-linear isometry, which is necessarily an $\mathbb R$-linear isometry because the norms are the same.
So $U(n)\subset O(2n).$
The groups $O(2n)$ and $U(n)$ act transitively on the unit sphere so define unique invariant probability measures $\nu_{O(2n)}$ and $\nu_{U(n)}$ respectively on $S^{2n-1}.$
The measure $\nu_{O(2n)}$ is $U(n)$-invariant because $U(n)\subset O(2n).$
By uniqueness, $\nu_{O(2n)}=\nu_{U(n)}.$
Let $\lambda^{2n-1}$ denote $(2n-1)$-dimensional Lebesgue measure on $S^{2n-1}$ and set $A_n=\lambda^{n-1}(S^{2n-1}).$
Then $\tfrac{1}{A_n}\lambda^{2n-1}$ is a $O(2n)$-invariant probability measure on the unit sphere, so is equal to $\nu_{O(2n)}.$

For completeness, here an argument for the unique existence of $G$-invariant probability measure on $M.$
I will write $\mu_G$ for Haar measure on $G,$ and $gm$ for the result of $g\in G$ acting on $m\in M.$
For any continuous $f:M\to\mathbb R$ define $\overline f(x)=\int f(g^{-1}x)d\mu_G(g).$ For any $x,y\in G$ there is some $h\in G$ with $hx=y,$ which gives $\overline  f(x)=\int f(g^{-1}h^{-1}y)d\mu_G(g),$ which is $\overline  f(y)$ by (left-)invariance of $\mu_G.$ So $\overline f$ is a constant function.
By the Riesz representation theorem, a measure $\nu$ on $M$ is determined by its integrals $\int f(x)d\nu(x).$For existence, pick any $m_0$ and define a probability measure $\nu$ on $M$ by $\int f(x)d\nu(x)=\overline  f(m_0)$ (using the Riesz representation theorem). For uniqueness, if $\nu$ is any $G$-invariant probability measure on $M$ then
\begin{align*}
\overline  f(m_0)&=\int \overline f(x)d\nu(x)&&\text{ because $\overline f$ is constant}\\
&=\int\int f(g^{-1}x)d\mu_G(g)d\nu(x)&&\text{ by definition of $\overline f$}\\
&=\int\int f(g^{-1}x)d\nu(x)d\mu_G(g)&&\text{ by Fubini}\\
&=\int\int f(x)d\nu(x)d\mu_G(g)&&\text{ by $G$-invariance of $\nu$}\\
&=\int f(x)d\nu(x)&&\text{ because the integrand did not depend on $g.$}
\end{align*}
$\Box$
