Image of a normal subgroup under automorphism is the same normal subgroup 
Let $G$ be a finite group. Let $H$ be a normal subgroup of $G$ such that the order of $H$ and the index of $H$ in $G$ are relatively prime. Let $f$ be an automorphism of $G$ and let $J = f(H)$. Prove that $J=H$.

Provided are two hints: 
Hint 1: Consider the orders of the subgroups $H\cap J$ and $HJ$. 
Hint 2: Consider the order of $\varphi(f(H))$ in $G/H$, where $\varphi: G \to G/H$ is the natural homomorphism.
I keep getting stuck when using the second hint as my approach. I feel like it is leading me towards using the First Isomorphism Theorem for groups but I'm having trouble implementing it.
 A: Using the second hint: By the hypotheses, the order of $H$ and the order of $G/H$ are relatively prime. Let's consider the cyclic subgroups of $f(H)$; their orders must divide $|H|$. The order of their images must divide both $|H|$ and $|G/H|$, and the only positive integer that does this is $1$. Hence every element of $f(H)$ is sent to the identity via the surjective homomorphism. Thus $f(H)\subseteq H$, hence by cardinality considerations we have $f(H)=H$.
A: To start, some words of a few things (propositions (1),(2)), from which it is easily follows answer. The subgroup $H$ of a finite group $G$ is a Hall subgroup, iff ${\rm gcd}(|H|,[G:H])=1$. If $H$ is a Hall subgroup, and $\pi=\pi(H)$, then $H$ is called $\pi$-Hall subgroup (not to be confused with Hall $\pi$-subgroup). 
(1) First show, that any $\pi$-Hall subgroup $H$ is a maximal $\pi$-subgroup. Let I - $\pi$-subgroup and $H\leq I$.
Denote $n=|G|$, $h=|H|$, $h'=[G:H]$, then $n=hh'$ and ${\rm gcd}(h,h')=1$. Since $I$ - $\pi$-subgroup, and $\pi=\pi(h)$, then ${\rm gcd}(|I|,h')=1$. By Lagrnge's theorem, $|I|\shortmid n=hh'$. So $|I|\shortmid h=|H|$. But $H\leq I$, hence $|H|\shortmid |I|$. Therefore, $|H|=|I|$ and $H=I$.
(2) Now let $H$ - normal $\pi$-Hall subgroup, and $I$ - any $\pi$-subgroup. We show, that $I\leq H$. Indeed, since $H\unlhd G$, then $HI\leq G$. Besides that,
$HI=\frac{|H||I|}{|H\cap I|}$, hence $HI$ - $\pi$-subgroup. Evidently, $H\leq HI$. By (1) $H=HI$. But $I\leq HI$, so $I\leq H$.
(3) Subgroup $H\leq G$ is called characteristic, if $f(H)=H$ for any automorphism $f$ of G. Show that any normal Hall subgroup $H$ characteristic. Let $\pi=\pi(H)$, $f$ - any automorphism of $G$, then $f(H)$ - $\pi$-subgroup of $G$, and by (2), $f(H)\leq H$. Since $f$ - automorphism, then $|f(H)|=|H|$, so $f(H)=H$.
A: Alternatively, observe that $N$ is generated by all $p$-Sylow subgroups of $G$, where $p \mid |N|$. In particular, $N$ is the unique subgroup of $G$ of order $|N|$.
A: If $|J|=|H|$, then $HJ$ (which is a subgroup by the normality assumption) has order 
$$ \frac{|H|\cdot|J|}{|H\cap J|}$$
If $|H\cap J| < |J|$, then there is some prime $p$ dividing $\frac{|J|}{|H\cap J|}$ . But then that means $p|H|$ divides $|HJ|$. Given that $p$ is a divisor of $|H|$, can you find a contradiction? What does that mean for $H\cap J$? How does this finish the problem?
