Show that if conjugation and left regular representations are equivalent then group is trivial

Question

Given that $G$ a finite group, we can define the conjugation and left multiplication representations $\rho_1$ and $\rho_2$ on $\mathbb{C}[G]$ respectively by:

$\rho_1(g)(r) = grg^{-1}$ and $\rho_2(g)(r) = gr$.

We want to show that if $\rho_i$ are equivalent then the group $G$ is trivial.

Attempt: Suppose equivalence then there is linear automorphism $\psi$ on $\mathbb{C}[G]$ such that $\psi(grg^{-1})=g\psi(r)$. I tried setting $r=e$ giving $\psi(e)=g\psi(e)$ and also $r\rightarrow rg$, giving $\psi(gr)=g\psi(rg)$. I know I should use $\psi$ is bijective somehow, but I don't see how to.

Answer 1

Okay, I found a way using character theory:

We have that for representations of finite groups over $\mathbb{C}$ the fact that they are equivalent if and only if their characters $\chi_1,\chi_2$ equal. Suppose $\exists g\in G$, $g\neq e$. Then since $\rho_1(g)(g)=g$ and $\rho_1(g)(e) = e$ the matrix of $\rho_1(g)$ (in the natural basis) has trace at least $2$. But $\rho_2(g)(h)=gh \neq h $ means $\rho_2(g)$ is trace-less. Contradiction. $\square$

Answer 2

Anytime a finite group $G$ acts on a finite set $X$ the multiplicity of the trivial representation inside the linearized representation $\mathbb{C}[X]$ is equal to the number of $G$-orbits of $X$. -- The indicator functions on each orbit are clearly $G$-invariant and linearly independent of with one another, and it's not too hard to see that in fact they form a basis for $\mathbb{C}[X]^G$.

There is one $G$-orbit of $G$ acting on itself by left multiplication (since every element is clearly in the orbit of the identity element as $g = g \cdot e$), and the $G$-orbits for the conjugation action are conjugacy classes.

Hence if these were isomorphic the multiplicities of the trivial representations would have to be the same, so their can only be one conjugacy class. But of course the identity element is always its own conjugacy class so this implies $G$ is trivial.

I'll also note that this works if we replace $\mathbb{C}$ by any commutative ring.