Non-modular implies class-determining

Statement
Suppose $$G$$ is a finite group and $$K$$ is a field whose characteristic is either zero or a prime not dividing the order of $$G$$. Then, $$K$$ is a fact about::class-determining field for $$G$$: given two representations $$\varphi_1,\varphi_2:G \to GL(n,K)$$ such that $$\varphi_1(g)$$ is conjugate to $$\varphi_2(g)$$ for every $$g \in G$$, it is true that $$\varphi_1$$ and $$\varphi_2$$ are equivalent linear representations, i.e., there exists a matrix $$A \in GL(n,K)$$ satisfying $$A\varphi_1(g)A^{-1} = \varphi_2(g)$$ for all $$g \in G$$.

Related facts

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

Facts used

 * 1) uses::Character determines representation in characteristic zero

Proof in characteristic zero
In characteristic zero, the proof follows directly from Fact (1), after noting that the character (trace) depends only on the conjugacy class of an element in $$GL(n,K)$$.

Proof in other characteristics
First, note that the characteristic zero proof does not work directly, because character does not determine representation in any prime characteristic.