I would like to find a unitary matrix $U$ which maps normalised vectors $\mathbf{v}^{(i)}$ from the set:
\begin{equation} \mathcal{V} = \Bigg\{ \begin{pmatrix} 1 \\ 0\\ 0 \\ \vdots \end{pmatrix}, \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1\\ 0 \\ \vdots \end{pmatrix}, \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 1\\ 1 \\ \vdots \end{pmatrix}, \ldots \Bigg\} \end{equation} into vectors $\mathbf{u}^{(i)} = U\mathbf{v}^{(i)}$ such that the magnitude of the elements of each $\mathbf{u}^{(i)}$ are equal i.e $|{\mathbf{u}^{(i)}}_j|= |{\mathbf{u}^{(k)}}_l| \,\, \forall \,\, i,j,k,l$.
If no such $U$ exists then is there a larger unitary which acts on the vectors $\mathbf{v}^{(i)}$ embedded in a larger vector-space as $\mathbf{v}^{(i)}\oplus \mathbf{0}$ ?
It is easy to see that if $\mathcal{V}= \big\{\begin{pmatrix} 1\\0\end{pmatrix}, \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1\\ \end{pmatrix} \big\}$ then $U =\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & i\\ i & 1 \end{pmatrix}$ satisfies the above criteria. Does this generalise to more than two dimensions? If not, I would like to know why.
I have tried to use various complex Hadamards to no avail. I've also tried using the cosine-sine decomposition to construct a unitary on $\mathbf{v}^{(i)}\oplus \mathbf{0}$ but with no success.
Thanks in advance, Patrick.
Update: I have been able to numerically find a three dimensional unitary which acts on $\mathbf{v}^{(1)},\,\mathbf{v}^{(2)}$ and $\mathbf{v}^{(3)}$ satisfying the criteria. I have tried numerically to solve the four dimensional case and have come very close but not solved it exactly (to numerical precision).
The only structure in the 3D case I observed is that the magnitude of the elements in the first column are all equal. This is obviously necessary in hindsight since $\mathbf{v}^1$ needs to be spread out uniformly.
