$p$ is the minimal prime dividing the order of $G$, and $H$ operates on $G/H$ by multiplication. $H/\ker\left(\varphi\right)$ embedded in $S_{p-1}$
Solution 1:
If a subgroup $H$ has minimal (prime) index, then it's a normal subgroup.
Moreover the action of $H$ on $G/H $ by left multiplication is given by
\begin{align} \ker(\varphi) & = \{h \in H| hgH = gH \ \text{for all}\ g \in G \} \\ & = \{h \in H| g^{-1}hgH = H \ \text{for all}\ g \in G \} \\\ \text{By normality} \ \ \ & = H \end{align}
So $\varphi$ is the trivial homomorphism.