Partition defined by orbits is trivial? So this may be a dumb question. I've stumbled across a potentially good false proof except I can't find the error. According to this reasoning, every partition defined by the orbits of the elements of G is the trivial partition (the whole of G):
So let the eq. relation $\sim$ be defined by $a \sim b$ iff $b \in orb(a)$. Obviously, the identity element $e$ is in every element's orbit, but if $a \sim b$ and $c \sim b$ then $a \sim c$ therefore, since every element in equivalent to $e$, all the elements are equivalent among themselves, therefore they are all in the same equivalence class. Another way of proving the same thing would be to say that in a group, every element can be 'made' from the combination of any other element with a specific element (ie the eq-n $a = xb$ always has a solution (trivially, $x = ab^{-1}$)), therefore the equivalence classes are all equal... my question is, where did I go wrong, if I did.

You say that $a\sim b $ iff $b\in orb(a)=\{ax| \ x \in X\}$ where $G$ acts on $X$.
So I suppose you have considered a very specific action, namely $G$ acts on $G$ by left multiplication. This action has indeed one orbit (i.e. one equivalence class)
