 Solution

If $\rho$ affords an irreducible character $\chi$, entrywise complex conjugation gives a representation with character $\overline{\chi }$. An invariant complex subspace for the conjugate representation conjugates back to one for $\rho$, so irreducibility is preserved. If representations afford $\chi ,\psi$, their tensor product has trace $\chi (g)\psi (g)$, proving that their product is a character.

With the usual character inner product,The character orthogonality and the first assertion therefore make this equal to one if $\chi =\overline{\psi }$ and zero otherwise. This also counts the invariant vectors in the tensor-product representation.