Let $n \geq r$ be two natural numbers and suppose $U \in \Bbb R^{n\times r}$ has orthonormal columns, i.e., $U^T U = I_r$. For any natural numbers $n \geq r$, is there a matrix $U \in \Bbb R^{n \times r}$ with orthonormal columns such that all rows of $U$ have the same $2$-norm, i.e., $\|v_1\|_2 = \dots = \|v_n\|_2$ where $v_i^T$ is the $i$-th row of $U$?
I know this holds trivially for $r=1$ and $r=n$. But I don't know how to prove it for general $r,n$. I tried to prove or disprove this for $n=3,r=2$, but I cannot construct such $U \in \Bbb R^{3\times 2}$.
For the $3 \times 2$ case you can try $$ \left[\begin{array}{cc} 0 & \frac{\sqrt{6}}{3} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{6}}{6} \\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{6}}{6} \end{array}\right] $$ Or for $4 \times 3$: $$ \left[\begin{array}{ccc} 0 & 0 & \frac{\sqrt{3}}{2} \\ 0 & \frac{\sqrt{6}}{3} & \frac{\sqrt{3}}{6} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{6}}{6} & -\frac{\sqrt{3}}{6} \\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{6}}{6} & \frac{\sqrt{3}}{6} \end{array}\right] $$ And $5 \times 4$: $$ \left[\begin{array}{cccc} 0 & 0 & 0 & \frac{2 \sqrt{5}}{5} \\ 0 & 0 & \frac{\sqrt{3}}{2} & -\frac{\sqrt{5}}{10} \\ 0 & \frac{\sqrt{6}}{3} & \frac{\sqrt{3}}{6} & \frac{\sqrt{5}}{10} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{6}}{6} & -\frac{\sqrt{3}}{6} & -\frac{\sqrt{5}}{10} \\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{6}}{6} & \frac{\sqrt{3}}{6} & \frac{\sqrt{5}}{10} \end{array}\right] $$ Hmm, seems to be a pattern...