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Not an answer to the whole question, but remarking on something that came up while thinking about it. If I think when $A$ is the a group ring of $G$, then according to the Brauer-Nesbitt theorem you just have it's not enough to check that the char polys on elements of $r_1(g)$ and $r_2(g)$ coincide for all $g \in G$. But that's not enough for buzzard's variantthe group, even in the case where $G$ is commutative -- you have to consider the whole algebra $A$. commutative/$\text{char}(k)=0$ case where buzzard can prove his Brauer-Nesbitt variant. Consider , for instance, the example where $G$ is an abelian group, $r_1$ is the trivial character, and $r_2$ is the sum of all the non-trivial characters (since every $g \in G$ is in the kernel of some nontrivial character).
Not an answer to the question, but remarking on something that came up while thinking about it. If $A$ is the group ring of $G$, then according to the Brauer-Nesbitt theorem you just have to check that the char polys of $r_1(g)$ and $r_2(g)$ coincide for all $g \in G$. But that's not enough for buzzard's variant, even in the case where $G$ is commutative -- you have to consider the whole algebra $A$. Consider, for instance, the example where $G$ is an abelian group, $r_1$ is the trivial character, and $r_2$ is the sum of all the non-trivial characters (since every $g \in G$ is in the kernel of some nontrivial character).