If $n_p=1$, then the $p$-Sylow subgroup is normal. Why? 
If $n_p=1$, then the $p$-Sylow subgroup is normal.

I've used this fact several times, and it's about time I knew why it's true.
 A: Let $N$ be a p-sylow subgroup.
For all $x$ ,$xNx^{-1}$ is a conjugate of $N$. If the set of groups conjugate to $N$ has size $1$ we find that $xNx^{-1}=N$ for all $x$.
A: I don't want to add something different here, but good to know that $n_p=[G:N_G(P)]$ when $P$ is a $p$- sylow of $G$. So if $n_p=1$ then $N_G(P)=G$ so $P$ is normal in $G$.
A: Well, $n_p$ is the number of $p$-Sylow subgroups. The conjugates of $p$-Sylow subgroup are precisely the other $p$-Sylow sugroups. So if $n_p=1$ then the $p$-Sylow subgroup is conjugate only to itself, i.e. it is normal.
A: Let $N$ be the single Sylow $p$-group. Suppose it is not normal. Then conjugation by some element $g$ gives a group $gNg^{-1}$ different from $N$ with the same number of elements as $N$. This is another Sylow $p$-group, distinct from the first, a contradiction. 
A: If I may add something.
The following statement is true:
Let $U\leq G$ be a subgroup such that there is no other subgroup that has the same order. Then $U$ is a characteristic subgroup. That is, every automorphism of $G$ maps $U$ to itself. 
