Def: Two representations $(V,\rho), (W,\tau)$ of $G$ are isomorphic if and only if there exists isomorphism $T:V\to W$ so $\forall g\in G, T\circ \rho_g=\tau_g\circ T$.
Let $G$ be a finite group, let $\rho$ be the regular representation. Let $\tau$ be the permutation representation induced by $G$ acts by conjugation on $G$. Is it true that $\rho$ and $\tau$ are isomorphic if and only if conjugation is a transitive action?
I'm able to show that the two representations are isomorphic then conjugation is transitive (by counting the size of orbits), but not the other way around. I failed to construct such a $T$ to make them isomorphic.
Best Answer
This happens at an earlier level than linear representations: if you have $G$-actions on $X,Y$ that are isomorphic then the associated permutation representations are isomorphic as well.
In particular if the conjugation action is transitive (I'll add some words about that at the end) then it is isomorphic to the regular action of $G$ on $G$ and thus the permutation representations are isomorphic.
However this is almost never the case : indeed for any $g\in G, geg^{-1} = e$ so if conjugation is transitive, $G=\{e\}$, which is not a very interesting scenario.