# Irreducible representations of a finite group are equivalent?

Given a finite group $$G$$. If $$D(g)$$ and $$D'(g)$$ are injective irreducible representations of $$G$$, do $$D(g)$$ and $$D'(g)$$ have to be equivalent? That is $$SD(g)S^{-1} = D'(g)$$?

We assume here complex irreducibility.

I don't think your statement is true. See irreducible representations of simple groups (say $$A_5$$) as examples: they are all injective if non-trivial.