= Solution
Schur <orthogonality> for irreducible complex characters of a <compact> <group>, with normalized <Haar measure>, is
$$
\langle\chi_\alpha,\chi_\beta\rangle:=\int_G\chi_\alpha(g)\overline{\chi_\beta(g)}\,d\mu(g)=\delta_{\alpha\beta}.
$$
Equivalent representations clearly have equal characters, since trace is invariant under conjugation. Conversely complete reducibility gives $\chi_V=\sum_\alpha m_\alpha\chi_\alpha$, where the nonnegative <integer> multiplicity is recovered by $m_\alpha=\langle\chi_V,\chi_\alpha\rangle$. Equal characters thus give equal multiplicities of every irreducible constituent, hence \b[equivalent representations].
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