I have been trying to solve this problem:
Let $G$ be a simple group of order 60. Determine $n_3$ and $n_5$ (the number of 3-Sylow and 5-Sylow subgroups, respectively).
This is what I have so far. I am least confident about why $\ker \varphi \neq G$ and for the same reason why a nontrivial kernel implies $n_3$ cannot be four.
According to the restrictions created by Sylow's theorems, the only possibility for $n_5$ is 6 ($n_5 \mid 12$ and $n_5 \equiv 1 \mod 5$), noting that $n_5$ as well as $n_3$ cannot equal 1 since $G$ is simple. The possibilities for $n_3$ are 4 and 10.
I then suppose that $n_3 = 4$ and try to come to a contradiction. There is the homomorphism $\varphi: G \to S_4$ induced by the group action under conjugation. If $\ker \varphi$ is nontrivial, then $n_3$ must equal 10.
If $\ker \varphi = e$, then $g \cong \varphi(S_4)$. But this can't happen since $60 = |G| > |S_4| = 24$.
If $\ker \varphi = G$, then all $P \in \text{Syl}_3(G)$ are fixed by $G$ in the homomorphism above (and therefore in the group action by conjugation). But this can't be since conjugation under the set $\text{Syl}_3(G)$ is transitive.
So, there is a nontrivial kernel. This means that there is some nonidentity $g \in G$ that fixes all $P \in \text{Syl}_3(G)$, but again, conjugation under this set is transitive. Then $n_3 \neq 4 \implies n_3 = 10$.
(Basically a comment that got a bit too long):
The argument could be more concise: If $\mathcal P_q$ denotes the set of Sylow $q$-subgroups of $G$, then conjugation induces an action of $G$ on $\mathcal P_q$, or equivalently a homomorphism $\phi_q \colon G \to \mathrm{Sym}(\mathcal P_q)$.
The Sylow theorems show that
Now if $\ker(\phi_q)=G$ then point 2. implies that $|\mathcal P_q|=1$ and so if $Q$ denotes the unique Sylow $q$-subgroup, $Q$ is normal in $G$. But if $G$ is simple, $1<|Q|$ and $Q$ normal implies $Q=G$, which is impossible since any $q$-group is solvable. Thus $\ker(\phi)$ is a proper normal subgroup of $G$ and hence $\ker(\phi_q) =\{e\}$ so that $\phi_q$ is an embedding.
If $G$ is simple and has order $60 = 2^2.3.5$ then $n_5=1 \mod 5$ and $n_5 \mid 12$ hence $n_5 \in \{1,6\}$, and hence $n_5=6$. On the other hand, $n_2 = 1 \mod 2$ and $n_2 \mid 15$, so that $n_2 \in \{1,3,5\}$. Since $60>3!$ it follows that $n_5=5$. Finally $n_3 =1 \mod 3$ and $n_4 \mid 20$ so that $n_3 \in\{1,4,10\}$ and since $60>4!$ it follows that $n_3 = 10$.
[Note that in the above we did not use anything about $G$ apart from its order, and, since $n_2=5$, we deduced that there is an embedding of $G$ into $S_5$, which thus has image $A_5$ so that $G \cong A_5$, i.e. up to isomorphism there is at most 1 group of order 60, and hence by checking $A_5$ is simple we can deduce that it is the unique simple group of order $60$. Thee simplicity of $A_5$ is easy to verify, for example, by considering its conjugacy classes.]