Let $A$ and $B$ in $O_n(\mathbb{R})$. Show that $A$ and $B$ commute. 
Let $A$ and $B$ in $O_n(\mathbb{R})$ (orthogonal matrices) such that $|||B-I_n|||<\sqrt{2}$ (subordinate norm) and $A$ commute with $BAB^{-1}$.
Show that $A$ and $B$ commute.

My 'attempt':
I know that $B^{-1}=B^{T}.$
We have $$ABAB^{T}=BAB^{T}A.$$
Since $A,B \in On(\mathbb{R})$ then $AA^{T}=I_n$ and $BB^{T}=I_n$.
Unfortunately I do not see how can I use the fact that $|||B-I_n|||<\sqrt{2}$.
Thank you in advance for your help.
 A: If $A$ and $BAB^{-1}$ commute, then they are diagonalizable (over $\mathbb{C}$) in a same orthonormal basis; let $$\mathbb{R}^n = E_1 \overset{\bot}{\oplus} \cdots \overset{\bot}{\oplus} E_r$$ be the associated decomposition. 
If $e_i \in E_i$ then there exists $\lambda_i \in \mathbb{C}$ such that $BAB^{-1}e_i= \lambda_i e_i$, hence $A(B^{-1}e_i)= \lambda_i (B^{-1}e_i)$. Therefore, $B^{-1}$ (and a fortiori $B$) permutes the eigenspaces of $A$, that is the $E_i$'s. 
Suppose by contradiction that there exist $i \neq j$ such that $BE_i = E_j$. Then for any $e_i \in E_i$ satisfying $\|e_i \|=1$,
$$2= \| Be_i \|^2+ \| e_i \|^2 = \| Be_i-e_i \|^2 \leq \| B- \operatorname{Id} \|^2< 2,$$
a contradiction. Therefore, $BE_i=E_i$ for every $1 \leq i \leq r$. Now, it can be deduced that $A$ and $B$ commute.
A: Unfortunately I can't give you solid proofs for the following, so this is really just a series of (hopefully) educated guesses....but I suspect the following, after a bit of investigation:


*

*if $\|B-I_n\|<\sqrt{2}$ then $B \in $SO$(n)$, so in other words $B$ is a "rotation". On dimensions 2 and 3 this certainly seems to hold...you can test it on a rotation matrix, and you will also find if you exchange columns so that the transformation is not a rotation you will always get $\|B-I_n\|>\sqrt{2}$. but I am not sure how this generalizes to higher dimensions...

*the proof will be complete if you can show that $BAB^T=A$...

*I'm guessing the fact that $A$ commutes with $BAB^T$ indicates that it is also in the group SO$(n)$, since SO$(n)$ is also a lie algebra, but I am not sure how this will help you to derive the result.


again some of this may be completely wrong, but could maybe help you to start looking in the right places for answers...   
