Can $g\cdot A < A$ for some group action?

Question

Can $g\cdot A$ be a strict subset of $A$ for some group action '$\cdot$' from $G\times G$ to $G$. For some subset $A$ of $G$, for some element $g$ other than identity in $G$.

Where $g\cdot A = \{ g\cdot a \mid a \in A \}$.

Answer 1

Let $\mathbb Q^*$ be the multiplicative group of nonzero rational numbers and let $A=\mathbb Z\setminus\{0\}$. Then if we take $g=2$ we have that $g.A$ is a proper subset of $A$.

Answer 2

Here is how I'm interpreting your question:
Let $G$ be a group and $A$ be a set and $\cdot$ be a group action $G\times A \to A$.
Let $g \in G$. Is it possible that $gA$ is a proper subset of $A$?

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No, this cannot happen. Let $G$ be any group acting on some set $A$.
Let $g \in G$. It is clear that $gA \subset A$.
To show the other inclusion, let $a \in A$ be arbitrary.
Note that $a = g\cdot\underbrace{(g^{-1}\cdot a)}_{\in A} \in gA$.

This shows the other inclusion as well.