This is an extension of my previous question on this topic. This time, we let $G=\underbrace{\mathbb{Z}/p^{k}\mathbb{Z} \times \ldots \times \mathbb{Z}/p^{k}\mathbb{Z}}_{n}$ and consider a subgroup of this $W \leq G$ such that for $W = \{(a_{1},a_{2},\ldots,a_{n}) : a_{i} \in \mathbb{Z}/p^{k}\mathbb{Z}, \sum_{i=1}^{n} a_{n} = 0 \}$. We know that $\text{Sym }(n)$ acts on on $G$ by permuting its factors. That is,
$$(a_{1},\ldots,a_{n})\sigma = (a_{\sigma(1)},\ldots,a_{\sigma(n)}).$$
Now notice that $\sum_{i}^{n}a_{i} = \sum_{i}^{n}a_{\sigma(i)} = 0$, so Sym$(n)$ will still act on $W$. But $W$can be expressed as:
$$\{(a_{1},\ldots,a_{n-1},-(a_{1} + a_{2} +\ldots + a_{n-1} ) ): a_{i} \in \mathbb{Z}/p^{k}\mathbb{Z}, \sum_{i=1}^{n} a_{n} = 0 \}$$
which i believe is isomorphic to
$$\{(a_{1},\ldots,a_{n-1}): a_{i} \in \mathbb{Z}/p^{k}\mathbb{Z} \}$$
so does this imply that, given that $\sigma$ can be viewed as a homomorphism and is bijective, that $\text{sym}(n) \leq \text{Aut}(\underbrace{\mathbb{Z}/p^{k}\mathbb{Z} \times \ldots \times \mathbb{Z}/p^{k}\mathbb{Z}}_{n-1})$
If $n>2$, then no permutation mapping $1$ to $2$ fixes $(1,0,\ldots,0,-1)$, but there's nothing special about $1$ and $2$, so no non-identity permutation acts trivially on $W$.
If $n=2$, then $(1,2)$ maps $(1,-1)$ to $(-1,1)$, so the action is nontrivial except when $p^k=2$.