The real $2$-dimensional irreducible representation of $C_3$ has endomorphism ring $\mathbb{C}$

A direct multiplication gives $A^2+A+I_2=0$, hence $A^3=I_2$. Therefore $g\mapsto A$ defines a real representation of $C_3=⟨g⟩$ on $V={\mathbb{R}}^2$.

Let $T=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. The condition $TA=AT$ from [L2] is equivalent to $c=-b$ and $d=a$, so the commuting endomorphisms are exactly the matrices $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$ with $a,b\in \mathbb{R}$.

Let $W\subseteq V$ be a nonzero invariant line, and choose $0\mathrm{\ne }v\in W$. Then $Av=\lambda v$ for some $\lambda \in \mathbb{R}$. Applying the polynomial identity from step 1.1 gives $({\lambda }^2+\lambda +1)v=0$, so ${\lambda }^2+\lambda +1=0$. But $4({\lambda }^2+\lambda +1)=(2\lambda +1{)}^2+3>0$ by [L3], impossible. Hence no nonzero proper invariant line exists, so the representation is irreducible by [L1].

The map $\mathrm{\Phi }:\mathbb{C}\to {\mathrm{End}}_{C_3}(V)$ given by $\mathrm{\Phi }(a+bi)=\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$ is bijective by the uniqueness in [L4], and the multiplication formula in [L4] matches matrix multiplication of these $2\times 2$ matrices. Therefore ${\mathrm{End}}_{C_3}(V)$ is a copy of $\mathbb{C}$.