Vector addition over orthogonal group I am working on a fun problem. The problem is to solve the following equation: $O_1x_1+O_2x_2=x_1+x_2$, where $x_1, x_2 \in \mathbb{R}^2$ are known and $O_1,O_2 \in \mathbb{O}(2)$ are unknowns. [$\mathbb{O}(2)$ is the group of orthogonal matrices.]
Also, consider the fact that $x_1 \neq \alpha x_2$ where $\alpha \in \mathbb{R}$.
My guess is that the solution is:
$$O_1x_1=x_1 \text{ and } O_2x_2=x_2$$
or,
$$O_1x_1=x_2 \text{ and } O_2x_2=x_1 \text{ if } \lVert x_1 \rVert = \lVert x_2 \rVert.$$
My attempt is:
$$\lVert O_1x_1+O_2x_2 \rVert^2= \lVert x_1+x_2 \rVert^2 \implies O_2^{\top}O_1x_1=x_1+\alpha x_2^{\perp}.$$
For any vector $d$, the following relationship holds, $$\langle {O_1x_1+O_2x_2 , d} \rangle = \langle {x_1+x_2 , d} \rangle .$$
I could not proceed further after this.
If my guess is correct, then I want an algebraical solution for this. If not, then I need an example to contradict it.
 A: Here is a counterexample.
$$ \left( \begin{array}{cc}
      0 & -1 \\
      1 & 0 \end{array} \right) \left( \begin{array}{c}
                                       1 \\
                                       0 \end{array} \right) + \left( \begin{array}{cc}
      24/25 & 7/25 \\
      -7/25 & 24/25 \end{array} \right) \left( \begin{array}{c}
                                       3 \\
                                       4 \end{array} \right) =  \left( \begin{array}{c}
                                       0 \\
                                       1 \end{array} \right) +   \left( \begin{array}{c}
                                       4 \\
                                       3 \end{array} \right) = \left( \begin{array}{c}
                                       1 \\
                                       0 \end{array} \right) +   \left( \begin{array}{c}
                                       3 \\
                                       4 \end{array} \right) $$
It clearly satisfies all the conditions. How did I find it? I focused on finding a counterexample with matrices in SO$(2)$, since SO$(2)$ is isomorphic to U$(1)$, that is, the complex numbers of norm $1$. A convenient isomorphism is given by:
$$ \left( \begin{array}{cc}
      a & b \\
      -b & a \end{array} \right) \mapsto a-bi $$
With this isomorphism it is easy to check that 
$$ \left( \begin{array}{cc}
      a & b \\
      -b & a \end{array} \right) \left( \begin{array}{c}
                                       c \\
                                       d \end{array} \right) = \left( \begin{array}{c}
                                       x \\
                                       y \end{array} \right) \Leftrightarrow (a-bi)(c+di)=x+yi $$
And so the question can be rephrased with complex numbers as having $ \alpha x_1 + \beta x_2 = x_1 + x_2$ with $\alpha$ and $\beta$ complex numbers of norm $1$ and $x_1$ and $x_2$ complex numbers which do not lie on a line in the complex plane. I set up this problem with $x_1 = 1$ and $x_2 = 3+4i$ and found by simple manipulation that $\alpha=i$ and $\beta = 24/25 - 7i/25$, which is the solution I gave above. 
A: Since you are working in $\mathbb{R}^2$, the answer can be seen geometrically: The orthogonal matrices consist of all rotations and all reflections. Furthermore, we know that the dot product (and hence the length of $x+y$) is conserved for all rotations. 
First separate angular and length part:
$O_1x_1+O_2x_2=O_1(x_1+O_1^TO_2x_2)$. Since $O_1$ doesn't change the length, you first have to answer for what orthogonal matrices will $x_1+Ox_2$ have the same length as $x_1+x_2$? Well
$$\langle x_1+Ox_2,x_1+Ox_2\rangle=\langle x_1,x_1\rangle + \langle x_2,x_2\rangle+2\langle x_1,Ox_2\rangle$$
But then, $\langle x_1,Ox_2\rangle=\langle x_1,x_2\rangle$, which can only be true for the identity and the reflection along $x_1$, since if $O$ is a rotation, it changes the angle between $x_1$ or $x_2$ and hence the scalar product. 
First suppose that $O$ was the identity. Then you only have to solve $\tilde{O}(x_1+x_2)=x_1+x_2$. There is only one rotation which can do that (identity) and one reflection (reflection along $x_1+x_2$). 
Now suppose that $O$ was the reflection along $x_1$. Then you have to solve
$\tilde{O}(x_1+Ox_2)=x_1+x_2$. Since orthogonal transformations are are rotations and reflections, we have two possibilites:


*

*$\tilde{O}$ reflects along an axis in between $x_1+x_2$ and $x_1+Ox_2$. This axis is clearly $x_1$, hence $\tilde{O}(x_1+Ox_2)=x_1+x_2$ and we have the identity.

*$\tilde{O}$ rotates $x_1+Ox_2$ to $x_1+x_2$. From a picture, you can then see that the result will be a reflection along $x_1+x_2$. If you don't like that, count angles: The angle between $x_1+Ox_2$ and $x_1+x_2$ is twice the angle between $x_1$ and $x_1+x_2$ since we did a reflection along $x_1$. Thus in total, we rotate $x_1$ by twice the angle between $x_1$ and $x_1+x_2$ in one direction (let's assume we rotated clockwise). We first reflect $x_2$ along $x_1$, which is equivalent to rotating it twice the angle between $x_1$ and $x_2$ (this time counterclockwise). Then we rotate it back twice the angle between $x_1$ and $x_1+x_2$ (this time clockwise again), which means that we can just as well rotate $x_2$ by twice the angle between $x_1+x_2$ and $x_2$ (counterclockwise). But this is equivalent to a reflection along $x_1+x_2$!
