@Shahrooz: You mean if one takes $G=Q_8\times A$ for a suitable infinite finitely generated abelian group $A$, then $G$ may satisfy conditions given in my question? I think these such groups cannot be counterexample for my question, as the question has positive answer for supersolvable groups. Note that abelian polycyclic groups are exactly finitely generated abelian groups.

If $\Gamma$ is bipartite then both $\lambda$ and $-\lambda$ are eigenvalues of $\Gamma$.

@RDK: From where you know $x^{2^d}$ is non-trivial? Note that the relation $x^{2^{d+1}}=1$ cannot solely imply that the order of $x$ is $2^{d+1}$. The answer of Derek Holt can help you to understand the main difficulty.

@Yassine: No. I have no idea. You may contact with M. Ghoraishi who has recently graduated under myb supervision his PhD.

@Yassine: I do not believe the conjecture is valid for any nonabelian finite $p$-group. The conjecture might br valid in the case of 2-generated metabelian as a characterization result of Caranti and Scopolla for such groups exist in terms of automorphism.

@Yassine: Another point is that as you mentioned above a possible counterexample must have order at least $p^8$, in this case GAP cannot help as the small group library is limited to groups of order at most $p^6$. But for the case of 2-groups of orders 512 or 1024 I do not know as some experts have access to the classification of these groups.