Consider a Lie algebra representation $\rho\colon\mathfrak{so}(3)\rightarrow \mathfrak{so}(n)$ that is irreducible over the reals. When $n$ is odd or divisible by four, such a representation exists. Indeed, when $n$ is odd, then the complex irreducible $n$-representation passes to a real representation. When $n$ is divisible by four, the direct sum of two copies of the complex irreducible $\frac{n}{2}$-representations gives a real representation, which is irreducible over the reals.
I want to understand the subspace of $n\times n$ real matrices that commute with every matrix in the image of $\rho$. If $n$ is odd, by Schur's Lemma, the only matrices that commute with a complex irreducible representation are multiples of the identity, giving us our answer. Consider $n$ divisible by four. While irreducible over the reals, $\rho$ is reducible over the complex numbers, so we cannot use Schur's Lemma.
Let $X$ be a matrix that commutes with every matrix in the image of $\rho$. Then the representation $\rho\otimes\rho^\ast$ acts trivially on $X$. When $n$ is odd, the decomposition of $\rho\otimes \rho^\ast$ contains one trivial summand, which Schur's Lemma tells us is the span of the identity matrix. When $n$ is divisible by four, we have $\rho=\underline{\frac{n}{2}}\oplus \underline{\frac{n}{2}}$, where $\underline{\frac{n}{2}}$ is the complex irreducible $\frac{n}{2}$-representation. Then $$\rho\otimes \rho^\ast=\left(\underline{\frac{n}{2}}\otimes\underline{\frac{n}{2}}\right)^{\oplus 4}.$$ This gives us four trivial summands. Here again, the identity matrix commutes with the representation, leaving us with three trivial summands.
I have computed the set of matrices commuting with $\rho$ when $n=4,8,12$. In each case, the space is contained in $\mathrm{span}(I_n)\oplus \mathfrak{so}(n)$. That is, if $X$ is traceless and commutes with $\rho$, then $X$ is antisymmetric. I was wondering if there was a representation theoretic reason for this phenomenon, coming perhaps from a version of Schur's Lemma over the reals.