I am reading Linear representations of finite groups by Serre, the following makes sense but can I see this with a concrete example since I cannot think of one?
Let $\rho$ and $\rho\,'$ be two representations of the same group $G$ in vector spaces $V$ and $V\,'$. These representations are said to be isomorphic if there exits a linear homomorphism $\kappa\colon V \rightarrow V\,'$ which ''converts'' $\rho$ to $\rho\,'$, if the following equality holds $$ \kappa \circ \rho(s) = \rho(s)\,'\circ \kappa \quad \forall s \in G.$$
If $\rho$ and $\rho\,'$ are given in matrix form by $\Gamma_s$ and $\Gamma_s'$, this implies, there exists an invertible matrix $K$ such that the following is holds: $$ \Gamma_s = K^{-1} \cdot \Gamma_s' \cdot K \quad \forall s \in G. $$
Let the base field be $\mathbb{C}$, and let $V=V'=\mathbb{V}_2(\mathbb{C})$.
Let $G=\mathbb{Z}_3=\langle d \rangle$.
In matrix terms let $$ \rho:d\mapsto \begin{pmatrix} 0 & 1\\-1 & -1\end{pmatrix} $$ and $$ \rho' :d\mapsto \begin{pmatrix}\omega & 0\\ 0 & \omega^2 \end{pmatrix} $$ where $\omega$ is the cube root of unity.
You can find the matrix of $\kappa$ for yourself, it's the usual matrix of eigenvectors.