A map $f : sp(1) \to sp(1)$ which commutes with the adjoint representation is a multiple of the identity Let $f : \operatorname{sp}(1) \to \operatorname{sp}(1)$ be a linear map which commutes with the adjoint representation of $\operatorname{Sp}(1)$, i.e. $f(\operatorname{Ad}_g X) = \operatorname{Ad}_g f(X)$ for all $g \in \operatorname{Sp}(1)$, $X \in \operatorname{sp}(1)$. I have managed to prove that necessarily $f = \lambda I$ by direct inspection (using the adjoint representation of $\operatorname{sp}(1)$ on itself, and the isomorphism $\operatorname{sp}(1) \cong (\mathbb{R}^3, \times)$).
However, the adjoint rep. of $\operatorname{Sp}(1)$ is irreducible, and that reminds me of Schur's lemma. It is not directly applicable, however, because $\operatorname{sp}(1)$ is a real vector space and not a complex one. Is it possible to argue in this way regardless? Perhaps by complexifying $\operatorname{sp}(1)$? I'm interested if there is some representation theory machinery that could squash this bug.
 A: Schur's lemma over a commutative ring $k$ says that the endomorphism $k$-algebra of an irreducible representation (more precisely, a simple object in any $k$-linear abelian category - the argument is very general) is a division algebra over $k$, which is finite-dimensional if the representation is finite-dimensional. When $k = \mathbb{R}$ this implies that the endomorphism algebra of a finite-dimensional irreducible representation is $\mathbb{R}$ or $\mathbb{C}$ or $\mathbb{H}$ (by the Frobenius theorem). But $\dim_{\mathbb{R}} \mathfrak{sp}(1) = 3$ is odd so it can't be a module over $\mathbb{C}$ or $\mathbb{H}$.
Using complexification will also work. It's a nice exercise to show that if $V$ is a representation over $\mathbb{R}$ then
$$\text{End}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}) \cong \text{End}_{\mathbb{R}}(V) \otimes_{\mathbb{R}} \mathbb{C}$$
and a computation of the complexifications of $\mathbb{C}$ and $\mathbb{H}$ will then show that a real irreducible representation has endomorphism algebra $\mathbb{R}$ iff its complexification is irreducible (provided that the complexification is semisimple - that’s true in this case and I think it’s true in general but it might require an additional argument), so you can try to show that. It amounts to showing that the adjoint representation of the complexification
$$\mathfrak{sp}(1) \otimes_{\mathbb{R}} \mathbb{C} \cong \mathfrak{sp}_2(\mathbb{C}) \cong \mathfrak{sl}_2(\mathbb{C})$$
is irreducible. You can check out Frobenius-Schur indicators for more on this sort of thing.
