If you have two finite-dimensional complex representations \(\pi_1\) and \(\pi_2\) of a finite group \(G\) such that \(\pi_1(g)\) is conjugate to \(\pi_2(g)\) for every g in \(G\), is it true that the two representations are isomorphic?
Solution
Yes.
1. The two representations have the same character.
For every \(g\in G\), the matrices \(\pi_1(g)\) and \(\pi_2(g)\) are conjugate. Conjugate matrices have the same trace, so \[ \chi_{\pi_1}(g)=\operatorname{tr}(\pi_1(g))=\operatorname{tr}(\pi_2(g))=\chi_{\pi_2}(g) \] for every \(g\in G\). Thus \[ \chi_{\pi_1}=\chi_{\pi_2}. \]
2. Finite-dimensional complex representations of a finite group are determined up to isomorphism by their characters.
By Maschke’s theorem both representations are completely reducible. Write \[ \pi_i\cong\bigoplus_{V\in\operatorname{Irr}(G)} m_{i,V}V. \] Taking characters gives \[ \chi_{\pi_i}=\sum_V m_{i,V}\chi_V. \] The irreducible complex characters are linearly independent (indeed orthonormal under the standard character inner product), so equality of the two characters forces \[ m_{1,V}=m_{2,V} \] for every irreducible \(V\).
3. Hence \(\pi_1\cong\pi_2\).
By 2, the two semisimple decompositions have the same irreducible multiplicities.