problem in group of order $2016$ 

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*Let $G$ a group of order $2016=2^5\cdot 3^2\cdot 7$ in which all elements of order $7$ are conjugate.
Prove that $G$ has a normal subgroup of index $2$.


I  know that all elements of order $7$ are conjugates of the same element. Let $n_7$ the number of Sylow $7-$ subgroups. Then if we consider the action of $G$ on $G$ by conjugation we have than
$$|\mathcal{O}(x)|\cdot|C_G(x)|=|G|$$ so $6n_7\cdot|C_G(x)|=|G|=2^5\cdot 3^2\cdot 7 \Rightarrow (7k+1)\cdot|C_G(x)|=2^4\cdot 3\cdot 7 \Rightarrow k=0$ or $k=1$.
If $k=0$ then $P_7\lhd G$ hence $|G/P_7|=2^5\cdot 3^2$ and it remains to find a subgroup of index $2$ in the quotient.
If $k=1$ then $n_7=8$ then we have $|C_G(x)|=2\cdot 3\cdot 7$ but can't see how to proceed either.
Question: Any ideas or hints on this problem?
[I found this in ALGEBRA QUALIFYING EXAM PROBLEMS
GROUP THEORY (Kent State University)]
 A: In the first case, you have an exact sequence $1 \rightarrow P_7 \rightarrow G \rightarrow G/P_7 \rightarrow 1$, hence a (by hypothesis transitive on $P_7 \backslash \{0\}$, because $P_7$ is cyclic) action of $G/P_7$ on $P_7$, ie a nontrivial homomorphism $h: G/P_7 \rightarrow Aut(P_7)=\mathbb{Z}/(6)$.
Note that any proper subgroup of $Aut(P_7)$ doesn’t act transitively on $P_7 \backslash \{0\}$, so that $h$ is onto. Then you’re more or less done.
In the second case, $G$ acts transitively by conjugation on its eight $7$-Sylow, so we get a morphism $h: G \rightarrow S_8$. If there is a $g \in G$ which acts as an odd permutation, we can consider the kernel of the signature and we are done – so assume it doesn’t happen and $h(G) \subset A_8$.
Take some $x$ of order $7$, consider its normalizer $N$ (of cardinality $2016/8=252$ by Sylow). Then, through $h$, $H$ is sent to a subgroup of $A_7$ (the group of even permutations that fix $x$).
If $N$ normalizes all $7$-Sylows, $x$ normalizes all $7$-Sylows so we have a map $\langle x \rangle \rightarrow Aut(P)$ for any $7$-Sylow $P$. But (cardinality) it follows that $x$ centralizes every $7$-Sylow. By conjugation every element of order $7$ centralizes $x$. But that makes $49$ elements centralizing $x$, a contradiction.
So $x \in N$ is mapped to a $7$-cycle of $A_7$.
But the centralizer $C_7$ and the normalizer $N_7$ of a $7$-cycle $c$ in $S_7$ can be shown to be in a short exact sequence $1 \rightarrow C_7 \rightarrow A_7 \rightarrow Aut(\langle c \rangle) \rightarrow 1$, and one easily sees that $C_7=\langle c \rangle =\mathbb{Z}/(7)$, so that $|N_7|=42$. Note that $(243756)$ normalizes $(1234567)$ without being in $A_7$, so that $N_7 \cap A_7$ has cardinality exactly $21$.
Thus $h(N)$ has cardinality dividing $21$, ie the kernel of $h$, intersected with $N$, has cardinality a multiple of $12$.
Thus the kernel $H$ of $G$ has cardinality a multiple of $12$ and isn’t divisible by $7$, $G’=G/H$ has cardinality dividing $168$ and a multiple of $7$.
Note that if $x,y \in G$ are of order $7$ and equal mod $H$, then $h(x)=h(y)$, so (fixed points) $x,y$ are in the same $7$-Sylow of $G$ thus are equal.
It follows that $G’$ has at least $48$ elements of order $7$, and as it has at most eight $7$-Sylows (of cardinality $7$) it has exactly $48$ such elements.
Thus $G \rightarrow G’$ is a bijection between the elements of order $7$, thus the action of $G’$ on its $48$ elements of order $7$ is transitive. But $48$ doesn’t divide $|G’|$ since $48$ doesn’t divide $168$...
