Let $\mathbb{M}$ the monster model of a theory $T$. If the orbit of an element $a$ over a set $A$ of elements reals, $O(a/A)$ is not finite then it must be of the size of $\mathbb{M}.$
In this case the orbit is related to the set $Aut_{A}(\mathbb{M})$ of automorphisms of $ \mathbb{M} $ that fixed the set $ A $, that is $$ O(a/A)= \{ b \in \mathbb{M}| \exists f \in Aut_{A}(\mathbb{M}), f(a)=b \} $$ as the monster model has the property that if $$ tp(a/A)=tp(b/A) \Rightarrow \exists f \in Aut_{A}(\mathbb{M}), f(a)=b $$ so I thought to try $tp(a/A)=tp(b/A)$ for all $b \in \mathbb{M}$ so for each $b \in \mathbb{M} \ \ \exists f \in Aut_{A}(\mathbb{M}), f(a)=b $ then $b \in O(a/A)$ and done. But I could not see this. Any hint is appreciated. I do not know if my idea is right or no. Thanks
Let $p(x)={\rm tp}(a/A)$ then $O(a/A)=p(\mathbb M)$. The following type is obviously not realized in $\mathbb M$
$q(x)=p(x)\cup \{b\neq x : b\models p(x)\}$
If $|p(\mathbb M)|<|\mathbb M|$, by saturation, it must be finitely inconsistent in $\mathbb M$. But $q(x)$ is finitely consistent whenever $p(\mathbb M)$ is infinite. Hence $O(a/A)=p(\mathbb M)$ is finite.