For any $n$, how can one characterize the set $S$ defined as
$$S=\{M\in \text{GL}(n,\mathbb{R})\text{ | }\forall A\in \text{SO}(n,\mathbb{R})\text{, }AM=MA\}$$
In other words, what does the space of $n\times n$ matrices which are invariant to conjugation by a rotation matrix look like?
Note $S$ is closed under addition, multiplication, inverse and scaling. Indeed, for any $M,N\in S$ and $A\in \text{SO}(n,\mathbb{R})$,
$A(M+N)=AM+AN=MA+NA=(M+N)A$
$A(MN)=(AM)N=M(AN)=(MN)A$
Letting $B=AM^{-1}$, we see $MB=(MA)M^{-1}=A(MM^{-1})=A\Rightarrow M^{-1}A=B=AM^{-1}$
For $\lambda\in \mathbb{R}$, we have $A(\lambda M)=\lambda (AM)=(\lambda M)A$
Furthermore, the identity matrix is clearly in $S$, trivially.
This is the set of matrices which commute with $C$, where $C$ is the vector subspace of $M_n(\Bbb R)$ generated by the elements of $SO(n)$. When $n=2$ then $$C=\left\{\pmatrix{a&b\\-b&a}:a,b\in\Bbb R\right\}.$$ The set of matrices commuting with $C$ is $C$, so $C=S$ here.
But for $n\ge3$, $C=M_n(\Bbb R)$ and the only matrices commuting with $C$ are the scalar multiples of $I$. If one can prove $C=M_n(\Bbb R)$ for $n=3$, then it's then easy to prove it holds for all higher values of $n$.