If the players are symmetric and the core is nonempty, then $x_i = v(N)/n$ for all $i$ is a Core element

Consider a TU game $$\langle N,v \rangle$$ with $$N = \{1,2,\ldots,n\}$$ being the set of players and $$v : 2^N \to \mathbb R$$, $$v(\emptyset) = 0$$ the characteristic function. The Core of $$v$$ is defined by undominated allocations $$x \in \mathbb R^n$$ \begin{align} \mathcal C(v) = \left\{x \in \mathbb R^n ~ \bigg| ~ \sum_{i \in S}{x_i} \geq v(S) ~ \forall S \subseteq N\right\}. \end{align} Let $$m_i(S) = v(S \cup \{i\}) - v(S)$$ denote the marginal contribution of $$i$$ to coalition $$S$$. We say that the players are symmetric if $$m_i(S) = m_j(S)$$ for all $$i,j \in N$$ and $$S \subset N$$. The marginal contribution of any agent to any coalition is thus identical.

I am wondering whether the following Proposition is true and how one proves it formally.

Proposition: If the players are symmetric and the core is nonempty $$\mathcal C(v) \neq \emptyset$$, then the equal division of the worth of the grand coalition is an element of the core $$(v(N)/n,\ldots,v(N)/n) \in \mathcal C(v)$$.

The center allocation $$z_{i}:=(v(N)/|N|)$$ for all $$i \in N$$ is in the core of the symmetric game $$v$$, whenever $$z(S) = \sum_{i \in S} z_{i} \ge v(S)$$ holds for all $$S \subseteq N$$. Now, if $$v(N)/|N| \ge v(S)/|S|$$ for all $$S \subseteq N$$, then it holds $$z(S) = \sum_{i \in S} z_{i} = |S|\cdot v(N)/|N| \ge v(S)$$. Thus, $$\mathbf{z} \in C(v)$$.
We observe that for a symmetric game $$v$$, the core is non-empty if $$v(N)/|N| \ge v(S)/S$$ holds for all $$S \subseteq N$$. Otherwise, the core is empty.