Centralizer containment is sufficient but not necessary

The averaging element $e=n^{-1}{\sum }_{j=0}^{n-1}{r}^j$ exists since $n$ is odd. Counting the $n$ occurrences of each $r^j$ in the square proves $e^2=e$. Conjugation by $r,s$ permutes its terms, so it is central in $kG$. Its ideal in $kH$ is the one-dimensional algebra $ke$, since $r^{j}e=e$, and augmentation sends $e$ to $1$. Hence $b_0(H)=ke$. F3 gives defect $1$, since $H$ has odd order.

The ideal $kGe$ has basis $e,se$, supported on the disjoint cosets $H,sH$. Its multiplication satisfies $(se{)}^2=e$, so it is $k{C}_2$. By F4 it has no nontrivial idempotents; consequently $e$ is primitive central in $kG$. As it acts by identity on the trivial module, $kGe=b_0(G)$. On restriction to $H\times H$ both $e$ and $se$ are fixed: $H$ is normal and $he=e$ for all $h\in H$. Thus the restriction is two copies of $ke$ as a bimodule.

For any other block idempotent $f$, $ef=0$, so left multiplication by $e$ is zero on the whole bimodule $kGf$ and its restriction. On $ke$ that operator is identity. Therefore $ke$ cannot be isomorphic to a summand of such a restriction. F1 proves uniqueness and $b_0(H{)}^G=b_0(G)$. Yet the centralizer of the trivial subgroup is $G$, while $H$ has index two. This contradicts the proposed necessity, without contradicting F2's sufficient implication. For $n=1$ the same computation is $H=1$, $G=C_2$; for the smallest nonabelian case $n=3$, $D_6\cong S_3\cong {\mathrm{S}\mathrm{L}}_2({\mathbf{F}}_2)$, the latter via its faithful action on the three nonzero vectors and its order six. No semisimplicity or choice assumption is needed.