Suppose $G$ is a finite group and $N$ a normal subgroup whose order $n$ is coprime to its index $m$. I start with an arbitrary section $\sigma: G/N \rightarrow G$. Then each $g \in G$ can be written as $g = n\sigma(\bar{g})$, where $n \in N$. Let $\ \Pi = N \times \underbrace{\ldots}_{m \text{ times}} \times N$ be the multiple direct product of $N$. Then there is a bijection between the sections $G/N \rightarrow G$ and $\Pi$. The action of $G$ on $\Pi$ is then defined by the right action of $G$ onto the sections. The orbits of $G$ can't all be of size $mn$ since the size of $\Pi$ is $m^n$. This means that there are sections whose stabilizer subgroup is non trivial. How can I find these sections, or am I wrong somewhere?
P.S. This problem is related to this question
As you are considering sections as sets, lets look at when an orbit is shorter, that is if an element $g$ stabilizes a particular section. That means that any section element $n_r\sigma(r)$ is mapped to another section element, that is $n_r\sigma(r)\cdot g=n_{r\cdot \bar g}\sigma(r\cdot \bar g)$, respectively $$ n_{r\cdot \bar g}=n_r\sigma(r)\cdot g/\sigma(r\cdot \bar g). $$ This implies that a section stabilizer may not have two different elements $g_1,g_2$ with same image in the factor group $\overline{g_1}=\overline{g_2}$, that is the candidates for section stabilizers are the subgroups that intersect trivial with $N$. Such subgroups act semiregularly on $G/N$ and the formula aboves prescribes section values with choices for each orbit of the subgroup.