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Let the group $G$ acting on a finite set $X$ with order $|X|=n$.

Because each element of $G$ permutes the set $X$, and there are $n!$ permutations of X, does this mean $|G|=|X|$?

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No, not at all in any way.

Remember that an action of a group $G$ on a set $X$ is just a homomorphism $G\to\operatorname{Sym}(X)$ from $G$ to the group of permutations of $X$. This homomorphism need neither be injective (so $|G|>n!$ is possible), nor surjective (so $|G|<n!$ is possible), and if the action is not transitive, we may well even have $|G|<n$.

In fact, any group can act on any set by means of the trivial action so that the mere existence of an action imposes no relation whatsoever between $G$ and $X$.

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If you add the hypothesis that $\phi : G \times X \rightarrow X$ is a $\textit{regular}$ action, i.e. that $$\forall x,y \in X. \ \exists!\ g \in G.\ y = \phi(g, x),$$ then you will have $|G|=|X|$.

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