Conclusions from Matrix multiplication commutation

Question

Let us suppose that for all matrices $B$ so that $B^{-1} = B^T$ we have $AB=BA$.

What conclusions if any, can we take from this?

For example can we say that $AB=BA=A$?

Thanks in advance.

EDIT:

To put it in another way:

What properties must $A$ have so that knowing $B$ is orthogonal, we can say that $AR=RA$?

For a $2 \times 2$ and a $3 \times 3$ matrix?

Answer 1

Take 3-dimensional rotations. If they commute it could mean that $A$ is rotation about the same axis as $B$.

Answer 2

No, we cannot say that $AB=BA=A$ for all such matrices. Take $A=I_n$ and $B$ different from $A$. Then $AB=BA$, but $AB=B\neq A$ for any $B$ with $B^{-1}=B^T$, different from $A$, say, for example $$ B=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \cr -1 & 1 \end{pmatrix}. $$