Let $G$ be a group. Let $G^* = \{\phi: G\to \mathbb{C}\}$ be the complex functions defined on $G$.
We have the representations
$$\rho: G \to GL(G^*), \space \rho(g) = \phi \mapsto (h\mapsto \phi(hg))$$
$$\psi : G \to GL(\mathbb{C}^{|G|}), \space \psi(g) = e_h \mapsto e_{gh}$$
right? (The latter is defined on the (standard) basis elements $e_h, h\in G$ and extended linearly).
I want to show that the linear isomorphism $\varphi: \mathbb{C}^{|G|} \to G^*$ that takes $e_h \mapsto \chi_{h}$ (characteristic function of $h$) is an isomorphism between the representations $\rho$ and $\psi$.
For this we must show that for a $e_h, h\in G$
$$\varphi(\psi(g) e_h) = \rho(g)(\varphi(e_h))$$
But $$\varphi(\psi(g) e_h) = \varphi(e_{gh}) = \chi_{gh}$$
and
$$\rho(g)(\varphi(e_h)) = \rho(g)\chi_h = \rho(g) (z\mapsto\chi_h(z)) = z\mapsto\chi_h(zg) = \chi_{hg^{-1}}$$
Where am I in the wrong?
PS. It takes me hours of pain in the neck to do these and usually they end up wrong like this. Each time I start from the beginning and do them mechanically with absolutely no understanding or routine. How to also achieve those?
You have two options:
Option 1: Define $\rho(g)$ by $\phi\mapsto (h\mapsto \phi(g^{-1}h))$. Then, $$ [\rho(g).\chi_h](x)=\chi_h(g^{-1}x)=\begin{cases}1&\mbox g^{-1}x=h\\0&\mbox{else.}\end{cases} $$ Therefore, $\rho(g).\chi_h=\chi_{gh}$ (since $g^{-1}x=h \iff x=gh$). Now, $$ \rho(g).\varphi(e_h)=\rho(g).\chi_h=\chi_{gh}=\varphi(e_{gh})=\varphi(\psi(g).e_h) $$ as desired.
Option 2: Define $\varphi(e_h)=\chi_{h^{-1}}$. Then, $$ \rho(g).\chi_h=\chi_{hg^{-1}} $$ as you observed above. Therefore, \begin{align} \rho(g).\phi(e_h)&=\rho(g).\chi_{h^{-1}}\\ &=\chi_{h^{-1}g^{-1}}\\ &=\chi_{(gh)^{-1}}\\ &=\varphi(e_{gh})\\ &=\varphi(\psi(g).e_h). \end{align}