= Solution
Existence of a minimal subgroup follows because $G$ has only finitely many subgroups and every module is relatively $G$-projective. Suppose an indecomposable module $M$ is relatively projective for both $H$ and $K$. Then $M$ is a summand of $\operatorname{Ind}_H^G U$ and of $\operatorname{Ind}_K^G V$ for suitable modules $U,V$. Applying the <Mackey restriction formula> and the <Krull–Schmidt theorem> shows that $M$ is relatively projective for some subgroup
$$
H\cap{}^gK.
$$
If $H$ and $K$ are minimal, this forces $H\subseteq{}^gK$. Reversing their roles gives the reverse containment after conjugacy; since the groups are finite, $H$ and $K$ are conjugate. Thus vertices form a unique conjugacy class.
Let $Q$ be a vertex and let $P$ be a Sylow p-subgroup of $Q$. Since $[Q:P]$ is invertible in $k$, every $kQ$-module is relatively $P$-projective. Transitivity of relative projectivity makes $M$ relatively $P$-projective, so minimality forces $Q=P$. Hence every vertex is a p-group.
For the trivial module $k$, every $kH$-endomorphism is scalar, and its relative trace to $G$ is multiplication by $[G:H]$. The <D. Higman criterion> says that $k$ is relatively $H$-projective exactly when $p\nmid[G:H]$. The minimal such subgroups are precisely the Sylow p-subgroups. Therefore the <vertex of an indecomposable module> gives
$$
\boxed{\text{the vertex of }k_G\text{ is a Sylow p-subgroup of }G.}
$$
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