Equivalence of group representations ($E(g)=S^{-1}D(g)S\text{for every }g\in G$)

Two finite-dimensional group representations $D,E$ over the same field are equivalent if there is one invertible linear map $S$ intertwining all their group actions, equivalently $E(g)=S^{-1}D(g)S$ for every $g$. The same change of basis must work simultaneously for all group elements. The cyclic property of the trace implies that equivalent representations have identical characters: $\mathrm{tr}(S^{-1}D(g)S)=\mathrm{tr}D(g)$. Consequently a difference at a single group element proves inequivalence. This implication does not need the group to be finite, and the reverse implication is not asserted for arbitrary groups or fields.