Cyclic implies every field is class-determining

Statement
If $$G$$ is a cyclic group and $$K$$ is a field, then $$K$$ is a class-determining field for $$G$$, i.e., given two representations $$\varphi_1, \varphi_2:G \to GL(n,K)$$ with the property that $$\varphi_1(g)$$ and $$\varphi_2(g)$$ are conjugate for every $$g \in G$$, it is true that $$\varphi_1$$ and $$\varphi_2$$ are equivalent linear representations, i.e., there exists a matrix $$A \in GL(n,K)$$ such that $$A\varphi_1(g)A^{-1} = \varphi_2(g)$$ for all $$g \in G$$.

Related facts

 * Elementary abelian of prime-square order implies corresponding prime field is not class-determining

Proof
The key idea is that we find the matrix that works for the generator of the cyclic group, and show that it works for the whole group.