Suppose to have a linear irreducible unitary representation $\rho:G\rightarrow U(H)$ on a complex Hilbert space $H$ with $G$ a generic group. Let $A$ be an $\textit{anti}$-linear operator such that 
$$
A\rho(g)=\rho(g)A\ \ \ \ \forall g\in G
$$
What can be said about the operator $A$? Does it hold anything like Schur's lemma?



Schur's lemma for antiunitary operators on complex Hilbert spaces
The part of Schur's lemma that continues to hold is that any such operator must be invertible or 0, if the representation is irreducible over the reals. I will make no assumption on complex (anti-)linearity from now on, but will assume that all operators are real linear.


The space of real operators commuting with $G$ must be a real division algebra, since it obviously contains the real numbers embedded as scalars. By the Frobenius theorem, there are only 3 possibilities for such algebras:


\begin{align\*}
&\mathbb{R}, \text{the real numbers}\\
&\mathbb{C}, \text{the complex numbers}\\
&\mathbb{H}, \text{the quaternions}\\
\end{align\*}
Indeed all of them are realized on real irreducible representations. The examples which admit anti-linear operators must be quaternionic. Amongst Lie groups, the smallest (non-finite) such example is $G=SU(2)$ acting on its 2-dimensional fundamental representation $\mathbb{C}^2$.


PS: Actually the 0-or-invertible property holds even if $H$ is not real-irreducible, since $A$ is a complex linear intertwining map between $H$ and the conjugate representation $\bar H$. It might not come from a division algebra in this case though.


