Show that there is an orthogonal matrix $O$ such that $OA_1=A_2O$. Let $A_1,A_2$ be two real $n \times n$ matrix. And suppose that they are two orthogonal and anti-symmetric matrices. Show that there is an orthogonal matrix $O$ such that $OA_1=A_2O$. 
I have no idea how to find such $O$.
 A: By assumption we have $A_i^2=-I_n$ for $i=1,2$, where $I_n$ is the identity matrix. In particular, $A_i$ is invertible. 
But any invertible anti-symmetric matrix is a symplectic form, that is, it must be similar to the canonical form
$$\begin{pmatrix} 0 & I_{n/2} \\ -I_{n/2} & 0 \end{pmatrix}.$$
Thus, $A_1$ and $A_2$ are similar. Note that $A_i$ is orthogonal, so we can choose the transition matrix to be orthogonal too. 
For example, choose any symplectic basis $e_1,\ldots,e_n$. (This is a standard result which can be done by induction.) Let's write $n=2m$. Then we have 
$$A_i(e_1,\ldots,e_m)=(e_{m+1},\ldots,e_{2m}), \quad A_i(e_{m+1},\ldots,e_{2m})=-(e_1,\ldots,e_m).$$
Apply Gram-Schmidt to $e_1,\ldots,e_m$. We have 
$$(e_1,\ldots,e_m)T=(e_1',\ldots,e_m'),\quad (e_{m+1},\ldots,e_{2m})T=(e_{m+1}',\ldots,e_{2m}').$$
Here $e_1',\ldots,e_m'$ are mutually orthogonal and of unit length. It is easy to show that $e_{m+1}',\ldots,e_{2m}'$ are also mutually orthogonal and of unit length, and that $e_1',\ldots,e_{2m}'$ is also a symplectic basis. Now under this new basis, which is orthonormal, $A_i$ has the desired matrix, and the transition matrix is orthogonal.
