Sets with small orbit

Question

Let ${\cal U}$ be some large saturated model and let ${\cal X}\subseteq{\cal U}$ be any set.

I would like to prove that the following are equivalent

1. $f[{\cal X}]={\cal X}$ for every automorphism that fixes a model containing $A$

2. $\big\{f[{\cal X}]\ :\ f\in\mathrm{Aut}({\cal U}/A)\big\}$ has small cardinality.

Edit. I accept @LevonHaykazyan's answer below though I was hoping in a more direct argument for 2$\Rightarrow$1. Morley/indiscernible/coheir sequences sound too much a sophisticated weaponry for such an innocent question.

Answer 1

1 implies 2. Let $M$ be a model of cardinality $\kappa$ containing $A$. Since $\mathcal X$ is $M$-invariant there is a set $P \subseteq S(M)$ of types such that $\mathcal X = \bigcup_{p \in P} p(\mathcal U)$. This means that the number of such $\mathcal X$ is bounded by $2^{2^\kappa}$. Now assume that the number of $A$-conjugates of $\mathcal X$ is unbounded. I claim that all of them are $M$-invariant which would be a contradiction. Let $f \in Aut(\mathcal U/A)$. Then $\mathcal X$ is $f^{-1}(M)$-invariant (since $f^{-1}(M)$ is a model containing $A$) and so $f(\mathcal X)$ is $M$-invariant. Indeed if $g \in Aut(\mathcal U/M)$, then $f^{-1}gf \in Aut(\mathcal U/f^{-1}(M))$ and so $f^{-1}gf(X) = X$ implying $gf(X) = f(X)$.

2 implies 1. Let $(\mathcal X_\alpha : \alpha < \lambda)$ be the set of $A$-conjugates of $X$. Define an equivalence relation $E(x, y) \leftrightarrow \bigwedge_{\alpha < \lambda} (x \in \mathcal X_\alpha \leftrightarrow y \in \mathcal X_\alpha)$. Then $E$ is $A$-invariant and has boundedly many equivalence classes. Each equivalence class of such an $E$ is $M$-invariant for any model $M$ containing $A$ (see e.g. Proposition 5.1 of Pierre Simon's book). This then shows that $\mathcal X$ is $M$-invariant since it is a union of equivalence classes.