Let $G$ be a finite group with a subgroup $H$ of index 5 and 5 is the smallest prime divisor of $|G|$.
Let $X=\{gH:g\in G\}$ be the set of left cosets of $H$ in $G$ and $H$ acts on $X$ by left multiplication.
Show that every orbit of $X$ has length 1.
I've tried by supposing that there is an orbit of length 2 but am stuck trying to show this and don't know if there is a better way?
Notice that $\{H\}$ is one orbit.
Consider an orbit $\Delta\ne\{H\}$ with $gH\in\Delta$ for some $g\in G$.
By the Orbit-Stabiliser Theorem $|\mathrm{Stab}_H(gH)|=|H|/|\Delta|$. In particular $|\Delta|$ divides $|H|$ and therefore $|G|$, but $|X|=5$ so $|\Delta|\le 4$.
Since $5$ is the smallest prime dividing $|G|$ this leaves $|\Delta|=1$ as required.