Representations Isomorphic up to a Character Suppose we have a finite group $G$ and with a normal subgroup $H$ such that the quotient is cyclic. Is it the case that two representations $\phi_1, \phi_2$ of $G$ are isomorphic when restricted to $H$ if and only if there is some 1-dimensional representation $\chi$ of $G$ so that $\phi_1 \otimes \chi \cong \phi_2$ and $\chi$ restricted to $H$ is trivial?
One direction is easy. For the other direction, I believe that $\chi$ (if it exists) should be of the form $Tr(\phi_1)Tr({\phi_2}^{-1})$, however I am having difficulty showing this is a homomorphism in the present case. Maybe there is an easier way to go about this?
Is this true in a more general context?
 A: Here's a partial answer: it's certainly true when $\phi_1$ and $\phi_2$ restrict irreducibly to $H$ (so they're both extensions of some irreducible representation of $H$). To see this, let $G$ act on $Hom_H(\phi_1,\phi_2)$ by $g\cdot f(v)=\phi_2(g)f(\phi_1(g^{-1})v)$. Restricted to $H$, this action is trivial ($f$ is $H$-equivariant), and you immediately see that $G$ acts on the hom space by a character of $G/H$, which is to say that $\phi_1=\phi_2\otimes\chi$ for some character $\chi$ of $G/H$. (And note that I didn't have to use anything about the quotient being cyclic here; this is true in generality).
If our irreducibility assumption isn't satisfied then we can't use Schur's lemma to see that the action is through a character. However, $G$ still acts on $Hom_H(\phi_1,\phi_2)$, which is some finite-dimensional space, i.e. $Hom_H(\phi_1,\phi_2)$ is a finite-dimensional representation of $G$, so that it splits as a direct sum of irreducible representations of $G$. These representations are still trivial on $H$, so they're irreducible representations of the cyclic group $G/H$, hence one-dimensional.
Write $d=\dim End_H(\phi_1)$, so that $G$ acts on $Hom_H(\phi_1,\phi_2)$ as $\chi_1\oplus\cdots\oplus\chi_d$, say. This means that, for all $f\in Hom_H(\phi_1,\phi_2)$, $g\in G$, $v\in V$, we have
$$\phi_2(g)f(\phi_1(g^{-1})v)=(\chi_1(g)+\cdots+\chi_d(g))f(v),$$
and so
$$f(\phi_1(g)v)=(\chi_1(g)+\cdots+\chi_d(g))\phi_2(g)f(v).$$
This just means that $f\in Hom_G(\phi_1,(\chi_1\oplus\cdots\oplus\chi_d)\otimes\phi_2)$, and so you can certainly say that one of the irreducible components of $\phi_1$ is isomorphic to a twist of one of the irreducible components of $\phi_2$; i.e. you're done if $\phi_1$ and $\phi_2$ are irreducible as representations of $G$.
If they aren't irreducible then I expect that there should be a counterexample, but I haven't really thought about constructing one.
