I have a two matrices $A$ and $B$ in $\mathbb{R}^{m \times n }$ ($m \gg $ n) such that there exists an orthonormal matrix $X \in \mathbb{R}^{n \times n }$, such that:

$$AX = B$$

Given that $X$ is orthonormal this is also true:

$$A = BX^T$$

**How to find $X$?**

I tried to use [Moore-Penrose inversion][1] $A^+$ and got non-orthonormal result $Y = A^+B$, that works only one way:

$$AY = B$$

but not in another. The problem is that both matrices $A$ and $B$ are not absolutely accurate (obtained in numerical calculations). So the non-orthonormal solution $Y$ arises. It is slightly more accurate than the exact solution $X$: $\|AY - B\| \lt \|AX - B\| < 0.001 $. But in the inverse case of course it doesn't work at all: $\|BY^T - A$$\| > 100.0$. Whilst the exact solution is good enough in both ways: $\|BX^T - A\| < 0.001$.

$A$ is full-rank, *i.e.* $A^+A = I$

So the question is how to find orthonormal solution of the overdetermined linear equations system?


  [1]: https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse#Linear_least-squares