Let $|G|=385=5\cdot 7\cdot 11$, and let $P$ be a Sylow $7$-subgroup of $G$. Then

The divisors of $55$ are $1,5,11,55$. Among these, only $1$ is congruent to $1$ modulo $7$. Hence

Therefore $P\trianglelefteq G$. Since $|P|=7$, $P$ is cyclic and

Conjugation gives a homomorphism

The image has order dividing both $385$ and $6$. Since

the image is trivial. Thus every element of $G$ centralizes $P$. Hence $P\leq Z(G)$. $\square$

Let $P$ be a Sylow $19$-subgroup of $G$. Then

The divisors of $55$ are $1,5,11,55$, and only $1$ is congruent to $1$ modulo $19$. Hence $n_{19}=1$, so $P\trianglelefteq G$. Since $|P|=19$, we have

The conjugation image has order dividing both $5\cdot 11\cdot 19$ and $18$. Their greatest common divisor is $1$. Therefore the conjugation image is trivial, and

Now let $Q$ be a Sylow $11$-subgroup. Then

The divisors of $95$ are $1,5,19,95$, and only $1$ is congruent to $1$ modulo $11$. Thus $n_{11}=1$. Hence $Q\trianglelefteq G$. $\square$

Let $|G|=100=2^2\cdot 5^2$. Let $P$ be the unique Sylow $2$-subgroup. Then $P\trianglelefteq G$ and $|P|=4$. Every group of order $4$ is abelian. Let $Q$ be a Sylow $5$-subgroup. Then

Therefore $n_5=1$, so $Q\trianglelefteq G$. Also $|Q|=25$, and every group of order $25$ is abelian. Since $P$ and $Q$ are normal subgroups of relatively prime orders, $P\cap Q=\{e\}$, and elements of $P$ commute with elements of $Q$. Moreover,

Thus $G=PQ$. Since $P$ and $Q$ are abelian and commute elementwise with each other, $G$ is abelian. Hence $G$ is abelian. $\square$

1. The Sylow $11$-subgroup is unique because $n_{11}\equiv 1 \pmod {11}$ and $n_{11}\mid 57$. The conjugation image has order dividing both $3\cdot 11\cdot 19$ and $10$, so it is trivial.
2. Since $n_{11}\equiv 1 \pmod {11}$ and $n_{11}\mid 95$, we get $n_{11}=1$.
3. A generator of $C_p$ can be sent to any of the $p-1$ nonidentity elements, and each such choice gives an automorphism.