If $H\unlhd G$ with $(|H|,[G:H])=1$ then $H$ is the unique such subgroup in $G$.

Alternatively, you might note that if $K$ is any other subgroup of order $|H|$, whether or not $K$ is normal (but assuming $H$ is normal), then $HK$ is a subgroup of $G$, so its order must divide $|G|$.

Let $L$ be any subgroup of $G$ with order $|H|$. Let $p$ be the natural projection $G \to \frac{G}{H}$. Then $p(L)$ is a quotient group of $L$, so $|p(L)|$ divides $|L|=|H|$. But $p(L)$ is also a subgroup of $\frac{G}{H}$, so $|p(L)|$ also divides $|G:H|$. By the coprimality hypothesis, $|p(L)|=1$ so $p$ is trivial on $L$. This means that $L \subseteq H$. Finally $L=H$ because those two subgroups have the same cardinality.

One way to see this is to look at the quotient map $f\colon G \to G/H$. Using Lagrange's theorem a couple of times, what can you say about the order of $f(K)$? [This isn't far from proving a well known formula for $\#(HK)$.]