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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.

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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.

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