Characters of elements under every representation equal implies conjugacy 
If $G$ is a group, suppose that for every $G$-module $V$ we have
  $$\chi_V(g_1)=\chi_V(g_2).$$ 
  How can I be sure $g_1$ and $g_2$ are conjugate in $G$?

Its easy to the reverse implication; that is $\chi_V(g)=\chi_V(hgh^{-1})\ \forall h\in G$. But how do I know $\chi_V$ won't map non-conjugates to the same element in $GL(V)$?
Thanks
 A: Here are counterexamples showing what happens when we drop any of the hypotheses listed by David Speyer in the comments. 
$G$ finite: The key point here is that characters are only defined for $V$ finite-dimensional, and it is possible to write down infinite groups $G$ which have no nontrivial finite-dimensional representations whatsoever. See, for example, this question. 
Ground field containing the $|G|^{th}$ roots of unity: Suppose we work over $\mathbb{R}$, and let $G = C_3 = \{ 1, \omega, \omega^2 \}$. Over $\mathbb{R}$, aside from the trivial representation there is only one other $2$-dimensional irreducible representation (coming from the embedding of $C_3$ into $\mathbb{C}^{\times}$), and this representation $V$ satisfies $\chi_V(\omega) = \chi_V(\omega^2)$. Maschke's theorem still holds here, so every representation is a direct sum of these, and it follows that characters don't separate $\omega$ and $\omega^2$. 
Ground field of characteristic zero: Suppose we work over a field of characteristic $3$, and let $G = C_3$ again. If $\rho$ is a finite-dimensional representation, then 
$$0 = 1 - \rho(\omega^3) = 1 - \rho(\omega)^3 = (1 - \rho(\omega))^3$$ 
hence $1 - \rho(\omega)$ is nilpotent, hence it has trace $0$. The same is true for $\rho(\omega^2)$, so characters don't separate any of the elements of $G$!
The first and third counterexamples both come, in some sense, from the failure of Maschke's theorem (more formally, the failure of the group algebra to be semisimple), whereas the second comes from the failure of representations to split. 
