Babai's problem

Let ${\displaystyle G}$ be a finite group, let ${\displaystyle \operatorname {Irr} (G)}$ be the set of all irreducible characters of ${\displaystyle G}$, let ${\displaystyle \Gamma =\operatorname {Cay} (G,S)}$ be the Cayley graph (or directed Cayley graph) corresponding to a generating subset ${\displaystyle S}$ of ${\displaystyle G\setminus \{1\}}$, and let ${\displaystyle \nu }$ be a positive integer. Is the set

${\displaystyle M_{\nu }^{S}=\left\{\sum _{s\in S}\chi (s)\;|\;\chi \in \operatorname {Irr} (G),\;\chi (1)=\nu \right\}}$

an invariant of the graph ${\displaystyle \Gamma }$? In other words, does ${\displaystyle \operatorname {Cay} (G,S)\cong \operatorname {Cay} (G,S')}$ imply that ${\displaystyle M_{\nu }^{S}=M_{\nu }^{S'}}$?

A finite group ${\displaystyle G}$ is called a BI-group (Babai Invariant group)[2] if ${\displaystyle \operatorname {Cay} (G,S)\cong \operatorname {Cay} (G,T)}$ for some inverse closed subsets ${\displaystyle S}$ and ${\displaystyle T}$ of ${\displaystyle G\setminus \{1\}}$ implies that ${\displaystyle M_{\nu }^{S}=M_{\nu }^{T}}$ for all positive integers ${\displaystyle \nu }$.

Which finite groups are BI-groups?[3]

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