= 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>,
$$
\langle\chi\psi,1_G\rangle=\frac1{|G|}\sum_g\chi(g)\psi(g)=\langle\chi,\overline\psi\rangle.
$$
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.
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