$\mathbf{Question:}$ Let $H, N$ be subgroups of a finite group $G$ with $N$ being normal in $G$. If the orders of $G/N$ and $H$ are relatively prime, then prove that $H$ is necessarily contained in $N$.
$\mathbf{Attempt:}$ Let $x \in H$ and let $o(G/N)=m$. Hence, $x^mN=N \implies x^m \in N$.
Now, $\gcd(m, o(H))$. Evidently $\phi:H \to H$ defined by $\phi(x)=x^m$ is an injection (as well as surjection) since $x^m=y^m \implies (y^{-1}x)^m=e \implies y=x $. Therefore, each element $h \in H$ can be written uniquely as $h=s^m$, for some $s \in H$.
Therefore, $\forall h\in H, \ h^m\in N \implies H\subset N$
Is this correct?
$\mathbf{Edit:}$ Changed $m \nmid o(H)$ to $\gcd(m, o(H))=1$
The argument you have used i.e. "$x^m=y^m \implies (y^{-1}x)^m=e $" is not (generally) correct.
The simplest proof is to note (Lagrange) that $|NH|$ is a factor of $|G|$.
However, $|NH|=|N||H|/|N\cap H|$ and so $|N|/|N\cap H|$ is a factor of $|G|/| H|$.
Since $|N|$ and $|G|/| H|$ are coprime the only possibility is that $|N|/|N\cap H|=1$.