Orbital maximin equals size of set for abelian groups

Statement
Suppose $$n$$ is a natural number greater than $$1$$. Then, for any abelian group $$G$$ acting on a set of size $$n$$, the maximum possible value of the minimum of the sizes of the orbitals is $$n$$. Further, there exists an abelian group (namely, the cyclic group of order $$n$$) for which this maximum is achieved.

This is a resolution to the fact about::orbital maximin problem for abelian groups.

Proof
We assume that $$n$$ is at least $$3$$. The result holds trivially for $$n = 2$$.

Reducing to the case of a transitive action
Suppose the action of $$G$$ on a set $$S$$ of size $$n$$ has more than one orbit. Let $$m$$ be the size of the smallest orbit, which we call $$\mathcal{O}$$. $$G$$ restricts to a permutation action on $$\mathcal{O}$$. In particular, the orbit of any orbital with both points in $$\mathcal{O}$$ lies completely inside the orbital set for $$\mathcal{O}$$.

Thus, if we prove the result for all $$m$$ for a transitive action, then using $$m \le n$$ would complete the proof. We can thus assume that the action is transitive.

Every element must act freely (semiregularly), i.e., without fixed points
We assume $$G$$ acts transitively on a set $$S$$ of size $$n$$.

Suppose we have an element $$\sigma$$ in $$G$$ and elements $$x,y,z \in S$$ such that $$\sigma(x) = x$$ and $$\sigma(y) = z$$. Since $$G$$ acts transitively, there exists $$\tau$$ such that $$\tau(x) = y$$. Then, $$\tau(\sigma(x)) = \tau(x) = y$$ whereas $$\sigma(\tau(x)) = \sigma(y) = z$$. Thus, $$\sigma \circ \tau \ne \tau \circ \sigma$$, contradicting abelianness.

Note that we can in fact deduce that the action must be a regular group action since it is both transitive and semiregular. This is part of a more general idea: any action that centralizes a transitive group action must be semiregular.

The size of an orbital is exactly $$n$$ for a transitive action
As before, $$G$$ acts transitively on a set $$S$$ of size $$n$$.

Consider any pair $$(x,y), x \ne y$$. We know that for any two elements $$\sigma_1, \sigma_2$$, if $$\sigma_1(x) = \sigma_2(x)$$, then $$\sigma_2^{-1}(\sigma_1(x)) = x$$. Since the action is free, this forces $$\sigma_1 = \sigma_2$$, also forcing $$\sigma_1(y) = \sigma_2(y)$$. Thus, no two distinct elements of the orbital of $$(x,y)$$ can have the same first coordinate. Since the number of first coordinates is $$n$$, we obtain that the size of the orbital is exactly $$n$$.