Vertices exist for indecomposable modules, are conjugate in G, and sources are conjugate by the appropriate normalizer

Choose a maximal $p$-subgroup $P\le G$. Then $[G:P]$ is prime to $p$, so the scalar $[G:P{]}^{-1}$ exists in $k$. Let $T$ be a left transversal for $G\mathrm{/}P$, and define $\alpha =[G:P{]}^{-1}{\mathrm{id}}_M\in {\mathrm{End}}_{kP}(M)$. By the relative trace formula in [L1], ${\mathrm{Tr}}_P^G(\alpha )(m)={\sum }_{t\in T}t\cdot \alpha (t^{-1}m)=[G:P][G:P{]}^{-1}m=m,$ so $M$ is relatively $P$-projective. Among the $p$-subgroups of $G$ for which $M$ is relatively projective, choose one minimal under inclusion and call it $Q$. Again by [L1], the adjunction counit from ${\mathrm{Ind}}_Q^G{\mathrm{Res}}_Q^{G}M$ to $M$ splits. Decompose ${\mathrm{Res}}_Q^{G}M$ into indecomposable summands using [L2]; one of their inductions must contain $M$ as a summand. That indecomposable summand is a source for $M$, so vertices and sources exist.

Let $R$ be another vertex. By [L1], choose $\alpha \in {\mathrm{End}}_{kQ}(M)$ and $\beta \in {\mathrm{End}}_{kR}(M)$ with ${\mathrm{id}}_M={\mathrm{Tr}}_Q^G(\alpha )={\mathrm{Tr}}_R^G(\beta )\mathrm{.}$ Expand the composite of these two trace expressions and group its terms by the double cosets $QxR$ of $Q\mathrm{\backslash }G\mathrm{/}R$. If $D_x=Q\cap {}^{x}R$, the terms in the $QxR$ block are permuted transitively by left conjugation from $G$ and their stabilizer is $D_x$; summing one set of stabilizer representatives gives a $D_x$-endomorphism ${\gamma }_x$. Thus direct regrouping of the finite double sum gives ${\mathrm{id}}_M={\mathrm{Tr}}_Q^G(\alpha ){\mathrm{Tr}}_R^G(\beta )={\sum }_{x\in Q\mathrm{\backslash }G\mathrm{/}R}{\mathrm{Tr}}_{D_x}^G({\gamma }_x)\mathrm{.}$ This is the needed Mackey trace calculation, with every summand now a $G$-endomorphism of $M$.

The endomorphism ring of the indecomposable finite-length $kG$-module $M$ is local by the Fitting argument in [L2]. Since the sum in step 1.2 is the identity, one summand $u={\mathrm{Tr}}_{D_x}^G({\gamma }_x)$ is invertible. Because $u^{-1}$ is $G$-linear, ${\mathrm{id}}_M=u{u}^{-1}={\mathrm{Tr}}_{D_x}^G({\gamma }_x{u}^{-1}),$ so [L1] makes $M$ relatively $D_x$-projective. Minimality of the vertex $Q$ forces $D_x=Q$, hence $Q\le {}^{x}R$. Interchanging $Q$ and $R$ gives $\mathrm{\mid }R\mathrm{\mid }\le \mathrm{\mid }Q\mathrm{\mid }$, while this containment gives $\mathrm{\mid }Q\mathrm{\mid }\le \mathrm{\mid }R\mathrm{\mid }$; consequently $Q={}^{x}R$. Thus vertices are conjugate in $G$.

Fix a vertex $Q$ and let $S,T$ be two sources attached to it. By [F1], $S$ is an indecomposable summand of ${\mathrm{Res}}_Q^{G}M$, and $M$ is a summand of ${\mathrm{Ind}}_Q^{G}T$. Hence $S$ is a summand of ${\mathrm{Res}}_Q^G{\mathrm{Ind}}_Q^{G}T\cong {⨁}_{x\in Q\mathrm{\backslash }G\mathrm{/}Q}{\mathrm{Ind}}_{Q\cap {}^{x}Q}^Q{\mathrm{Res}}_{Q\cap {}^{x}Q}^{{}^{x}Q}({}^{x}T),$ where the displayed decomposition follows by partitioning $G$ into its $Q$-$Q$ double cosets and grouping the corresponding induced-function summands. By Krull-Schmidt [L2], $S$ is a summand of one displayed term and is therefore relatively $Q\cap {}^{x}Q$-projective for some $x$.

The source $S$, viewed as a $kQ$-module, cannot be relatively projective for a proper subgroup $D<Q$: otherwise induction transitivity would make $M$, a summand of ${\mathrm{Ind}}_Q^{G}S$, relatively $D$-projective, contradicting minimality of the vertex $Q$. Applying this to step 2.2 forces $Q\cap {}^{x}Q=Q$. Equality of the finite subgroup orders then gives ${}^{x}Q=Q$, so $x\in N_G(Q)$. The corresponding summand in step 2.2 is ${}^{x}T$, which is indecomposable; since $S$ is an indecomposable direct summand of it, $S\cong {}^{x}T$. Thus sources attached to the fixed vertex are conjugate by an element of $N_G(Q)$.

Steps 1.1 through 3.1 prove the theorem, with no stronger uniqueness claim than the normalizer-conjugacy stated above.