Let $\,Q_p\,$ be a Sylow p-sbgp. of $\,G\,$ . If either $\,Q_5\,,\,Q_7\,$ is normal then the product $\,P:=Q_5Q_7\,$ is a sbgp. of $\,G\,$ and thus it is normal as well, as its index is the minimal prime dividing $\,|G|\,$. In this case, and since the only existing group of order $35$ is the cyclic one, we'd get that $\,|\operatorname{Aut}(P)|=\phi (35)=24$, and thus we can define a homomorphism $\,Q_3=\langle c\rangle\to\operatorname{Aut}(P=\langle d\rangle)\,$ by $\,d^c:= c^{-1}dc\,$ . As this homom. isn't trivial (of course, we assume at least one of the Sylow sbgps. is non-abelian) we get a (non-abelian, of course) semidirect product $$ P\rtimes Q_3 $$ .
If both $\,Q_5,Q_7\,$ are not normal, a simple counting argument tells us there are $\,90\,$ elements of order $7$ and $84$ elements of order 5, which is absurd as we've only $\,105\,$ elements in the group.
Thus, the above is the only way to get a non-abelian group of the wanted order.
Assume $G$ is simple and that $n_5=6$, so index$[G:N_G(P)]=6$, $P \in Syl_5(G)$. But then $G$ embeds homomorphically into $A_6$ (note that core$_G(N_G(P))=1$, since $G$ is simple!). But index$[A_6:G]=360/180=2$, implying $G$ is normal in $A_6$, which contradicts the simplicity of $A_6$.
Best Answer
The image of the conjugation action on the six Sylow $5$-subgroups is a perfect subgroup of $A_6$ with order divisible by $30$. It can't be order $180$ because that would be a subgroup of index $2$ in $A_6$. It can't be order $30$ or $90$, because they have twice odd order and hence have subgroups of index $2$.
So the image has order $60$ and the kernel $K$ has order $3$. Since $G$ is perfect, we must have $K \le Z(G)$.
But now we can use 10.1.6 of Robinson's book. A Sylow $3$-subgroup $P$ of $G$ is abelian, and since $G$ is perfect the transfer homorphism $\tau:G \to P$ is the trival map, but then $C_P(N_G(P)) = 1$, contradicting $K \le C_P(N_G(P))$.