If $G$ is finite and $H$ is the only subgroup of a given order, then $H$ is normal. 
If $G$ is finite and $H$ is the only subgroup of a given order, then $H$ is normal.

I have a proof idea that I'm not sure works or not:
By Lagrange's theorem, the order of $G$ is equal to the order of $H$ times the number of distinct left cosets of $H$. But every left coset of $H$ has the same number of elements as $H$, so the order of $H$ must be equivalent to the order of $G$. Hence, $H = G$, so $H$ is normal.
Does this work? I'm aware of the proof using the conjugation map, but was curious to know if this works or not. Thank you!
 A: This does not work, since cosets are not subgroups (except for the trivial cosets), and your conclusion is not true. For instance, if $G$ is a cyclic group of order $2n$, then it has a unique subgroup of order $n$ (which is of course normal, since the group is abelian, but it is also definitely not all of $G$ if $n>0$).
Instead, recall that $H\leq G$ is normal iff $H^g=H$ for all $g$.
A: The claim is correct (and in fact, can be made stronger: If a subgroup is the only one of its order then it must be characteristic). But your argument idea is not a valid proof if it, you get that $|H|$ divides $|G|$, but they are not generally equal.
Useful to know for this proof:
A subgroup is characteristic iff it is invariant (as a set) under every automorphism.
A subgroup is normal iff it is invariant (as a set) under every inner automorphism (or equivalent, every conjugation action $ghg^{-1}$ for $g\in G$ and $h\in H$)
You can easily argue that is it characteristic and therefore normal (hint, where else could an automorphism of $G$ take $H$ but back to itself? and why?)
