Could anyone please help me to construct an orthogonal matrix which would rotate the first basis vector $v_1$ from an orthonormal set $\{v_k\}_{k=1}^N$ into a vector having equal projections on all the basis vectors? $$ U v_1 = v_\Sigma = \dfrac{1}{\sqrt{N}}\sum_{k=1}^N v_k $$ I'm looking for the simplest possible rotation, the one which would go in the $(v_1,v_\Sigma)$ plane, without any additional extra rotations around $v_1$ or $v_\Sigma$.
Right away I can tell that $$ O_N = \begin{pmatrix} \frac{1}{\sqrt{N}} & \cdots & \cdots & \cdots \\ \frac{1}{\sqrt{N}} & \cdots & \cdots & \cdots \\ \cdots & \cdots & \cdots & \cdots \\ \frac{1}{\sqrt{N}} & \cdots & \cdots & \cdots \\ \end{pmatrix} \ . $$
How do I proceed further?
UPDATE
So the answer turns out to be:
$$
O_N = \begin{pmatrix}
\frac{1}{\sqrt{N}} & - \frac{1}{\sqrt{N}} & - \frac{1}{\sqrt{N}} & -\frac{1}{\sqrt{N}} & \cdots & - \frac{1}{\sqrt{N}}\\
\frac{1}{\sqrt{N}} & a & b & b & \cdots & b\\
\frac{1}{\sqrt{N}} & b & a & b & \cdots & b\\
\frac{1}{\sqrt{N}} & b & b & a & \cdots & b\\
\cdots & \cdots & \cdots & \cdots & \ddots & \cdots \\
\frac{1}{\sqrt{N}} & \cdots & \cdots & \cdots & \cdots & a\\
\end{pmatrix} \ ,
$$
where
$$
a = \dfrac{1}{1+\dfrac{1}{N+\sqrt{N}-1}} \ , \quad
b = - \dfrac{1}{N + \sqrt{N}}
$$
(The answer was obtained in a brute-force way, using Mathematica's functions RotationMatrix[] and FindSequenceFunction[].)
I'm not sure I follow the simplest possible rotation criterion. I assume this is an n-dimensional space. Here's a very easy solution:
consider the image of U on the entire orthonormal set / unitary matrix
$ \bigg[\begin{array}{c|c|c|c} U\mathbf v_1 & U\mathbf v_2 & \cdots & U\mathbf v_n \end{array}\bigg] = U V = V Q$
now select $Q$ to be the Discerete Fourier Transform, so
$U:=VQV^*$
gives a solution
note in particular
$UV\mathbf e_1 = U\mathbf v_1 = VQ\mathbf e_1 = \frac{1}{\sqrt{n}}V\mathbf 1=\frac{1}{\sqrt{n}} \sum_{k=1}^n \mathbf v_k$