It is easy to see that for any $x$ and $y$ on the unit sphere of a Hilbert space $H$ there exists a surjective isometry $U$ such that $Ux=y$. Does something more general also hold? That is, given two pairs $(x\_1, y\_1)$ and $(x\_2, y\_2)$ of vectors on the unit sphere such that $||x\_1-x\_2||=||y\_1-y\_2||$, can we find a surjective isometry $U$ such that $Ux\_1=y\_1$ and $U x\_2=y\_2$?



Isometries of Hilbert space
For complex scalars, this fails in $\mathbb{C}^2$. Take $x\_1 = y\_1 = (1,0)$, $x\_2 = (i,0)$, $y\_2 = (0,1)$. No (complex linear) isometry can take the one-dimensional subspace spanned by $x\_1$ and $x\_2$ onto the two-dimensional subspace spanned by $y\_1$ and $y\_2$.


For real scalars, it is true. It suffices to find an isometry from ${\rm span}(x\_1,x\_2)$ to ${\rm span}(y\_1,y\_2)$ which takes $x\_1$ to $y\_1$ and $x\_2$ to $y\_2$. This just means that you have two points on the unit circle in $\mathbb{R}^2$ and I have two points on the unit circle, and the distance between your two points equals the distance between my two points, and you want an isometry of $\mathbb{R}^2$ to itself that takes your points to mine. Which is easy.


