= Solution
A <group representation> assigns each $g\in G$ an invertible linear map $D(g)$ on a fixed vector space, preserving multiplication:
$$
D(gh)=D(g)D(h),\qquad D(e)=I.
$$
Choosing a basis turns those maps into matrices. A <faithful representation> is injective, equivalently its kernel is $\{e\}$, so distinct elements have distinct matrices.
Two <group representations> on spaces of the same dimension are called <equivalent representations> if one fixed invertible change of basis conjugates every representing matrix: $E(g)=S^{-1}D(g)S$ for all $g$. Choosing a different $S$ separately for each element does not establish equivalence.
The <character of a representation> is
$$
\boxed{\chi_D(g)=\operatorname{tr}D(g).}
$$
The <cyclic property of the trace> makes it invariant under a change of basis and constant on conjugacy classes. Thus different characters at even one group element prove inequivalence.
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