Image of a normal Hall Subgroup under an automorphism Let $G$ be a group such that $|G|= n = md$, where $\gcd(m,d)=1$. 
Let $N$ be a normal subgroup of $G$ with order $m$. 
Further, let us define a subgroup $H$ of order $d$. I managed to prove that, $H \cap N= [e] \implies HN = G$, and $HN \cong H \times N$.
Now, let $f$ be an automorphism of $G$.
Then, $|f(N)|= m$.
Consider that for some $n_i \in N$,
$f(n_i^m)= f(n_i)^m \implies f(e) = e = (f(n_i))^m$
So, the order of $f(n_i)$ divides $m$.
As $f(n_i) \in G$, $f(n_i) \in h_i N$, for some $h_i \in H$. So, let, $f(n_j) = h_i n_i \implies (h_in_i)^m = e$.
Now, consider a bijective map $g: HN \to H \times N$, such that,
$f(h_i n_i) = (h_i, n_i) \implies f((h_i n_i)^m)= f(e)= (h_i, n_i)^m= (h_i^m, n_i^m)$
$\implies (h_i^m, n_i^m)= (e, e) \implies h_i^m= e$ which is simply not possible given that $h_i \in H$ and that $gcd(m,d)=1$, unless, $h_i = e$.
So, for all $n_i \in N, f(n_i) \in N \implies f(N) = N$.
Okay, well, can somebody please just look through this reasoning and see if this is acceptable? Also, it seemed somewhat unnecessarily wordy. How could I prove it with greater brevity?
Thank you!
 A: There are several ways to do it. One is to note that
$$
\lvert N f(N) \rvert = \frac{\lvert N \rvert \cdot \lvert f(N) \rvert}{\lvert N \cap f(N) \rvert}.
$$
So this number is a multiple of $m$, a divisor of $n = m d$, and of $\lvert N \rvert \cdot \lvert f(N) \rvert = m^{2}$, and thus coprime to $d$. The only choice is 
$$
\lvert N f(N) \rvert = m = \lvert N \rvert,
$$
so $f(N) \subseteq N$, and thus $f(N) = N$.
Alternatively, I think you are proving that
$$\tag{eq}
N = \{ x \in G : x^{m} = 1 \},
$$
which immediately implies that $N$ is invariant under automorphisms (even endomorphisms).
(eq) follows from the fact that if $x^m = 1$, then $(x N)^m = N$ in $G/N$, but since the order of $G/N$ is $d$, you also have $(xN)^d = N$, and since $(m, d) = 1$, we have $x N = N$, that is, $x \in N$.

PS The second way of doing it proves the statement that @TobiasKildetoft made in his comment. The first way works too, just take any subgroup $M$ of order dividing $m$, and consider $\lvert N M \rvert$, to obtain $M \le N$.
