x'}=R {\bf x}$ and ${\bf y'}= R {\bf y}$ to prove Lemma $[1]$.

A transformation $R \in K(n)$ can be described as a symplectic map: $$\label{symplectic}
R = R(X,Y) \equiv
\begin{bmatrix}
X & Y  \\ 
-Y & X  \\ 
\end{bmatrix}$$ where the matrices $X$ and $Y$ are such that [@ADMS95]: $$\begin{aligned}
X^T X+Y^T Y &=& X X^T + Y Y^T =1\\
X^T Y &,& X Y^T \quad \mathrm{symmetric}.\end{aligned}$$ Note that this matrix is written for reordered vectors of the form $[x_1, x_3, \cdots, x_{2n-1},x_2, x_4, \cdots, x_{2n}]$, that is, one first writes the $n$ $q$-quadratures then the $n$ $p$-quadratures for all the vectors.

Let us introduce the following vectors ${\bf a}, {\bf a'}, {\bf b}, {\bf b'} \in \mathbb{C}^n$ defined as $$\begin{aligned}
a_k = x_{2k-1} + i x_{2k} &,& a'_k = x'_{2k-1}+ i x'_{2k}\\
b_k = y_{2k-1} + i y_{2k} &,& b'_k = y'_{2k-1}+ i y'_{2k}.\end{aligned}$$ Then, the conditions read: $$\label{condition2}
\left\{
\begin{array}{ccc}
||{\bf a}||^2=||{\bf a'}||^2\\
||{\bf b}||^2=||{\bf b'}||^2\\
\mathrm{Re}\langle {\bf a} | {\bf b} \rangle = \mathrm{Re}\langle {\bf a'} | {\bf b'} \rangle
\end{array}
\right. ,$$ where $\mathrm{Re}(x)$ refers to the real part of $x$.

For our purpose, it is therefore sufficient to prove that there exists an unitary 