Number of orbits of a finitely generated group

Question

Let $G$ be a group. Define an action $G \times G \to G: (x,y) \mapsto y^{-1}xy$ on $G$. Let $ x$ be an arbitrary element of $G$ an denote by Orb$(x)$=$\lbrace y^{-1}xy| y \in G \rbrace$ the orbit of $x$. Write Orb$(G)$ for the set of orbits of elements of $G$. If $G$ is finitely generated, say $G= \langle x_i| i \in I \rangle$ for a certain index set $I$ with $|I| < \infty$. Does this imply that Orb$(G)$=$\lbrace Orb(x_i)|i \in I \rbrace$?

Answer 1

No, take $G=\mathbb{Z}^2$, then $(1,0)$, $(0,1)$ are generators.

The orbit of $(1,1)$ is $\{(1,1)\}$ (because the group is Abelian).

The orbit of $(1,0)$ and $(0,1)$ is $\{(1,0)\}$ and $\{(0,1)\}$ respectively, as you can see non of them is equal to the orbit of $(1,1)$.