Let $(\pi,V)$ be a finite dimensional complex representation of a group $G$. If $G$ is finite, then $\pi$ is determined up to isomorphism by its character $\chi_{\pi}$.
What about the case where $G$ is not necessarily finite? I doubt it's still true that $\pi$ is determined by its character, even up to semisimplification.
If $\pi_1, \pi_2$ are two semisimple representations of $G$ with the same character, what other properties would $\pi_1$ and $\pi_2$ have to share to conclude they are isomorphic? For example, suppose $H$ is a subgroup of finite index of $G$, and we have not only $\chi_{\pi_1} = \chi_{\pi_2}$, but also $\pi_1|H \cong \pi_2|H$.
The reason I'm asking is I have a semisimple representation $\pi$ of a group $G$ which I believe is isomorphic to some induced representation $\operatorname{Ind}_H^G \sigma$ for a semisimple representation $\sigma$ of $H$. I know how to directly construct an isomorphism between these guys, but it's very tedious. But it's easy for me to show that their characters are equal and that their restrictions to $H$ are isomorphic.
The characters being equal is all you need. Even for infinite groups it is true that two semisimple representations are isomorphic iff they have the same character. Of course you can have non-semisimple representations with the same characters which are non-isomorphic though.