This statement is true for definable sets, but how about type-definable sets?
I encountered with this question when reading the proof of existence of $G^{00}$ for $\varnothing$-type-definable group $G$ in NIP theory. There, Pierre Simon assumed toward contradiction that $G^{00}$ does not exist and got the existence of an arbitrary large sequence of distinct type-definable subgroups of bounded index. Howeover, one step is actually missed here, i.e. we need to show that if there does not exist such an arbitrary large sequence, $G_A^{00}$ does not depend on $A$. To do this, we pick up the smallest type-definable subgroup $G_A^{00}$ over arbitrary parameter sets. Note that $G_{A}^{00}$ must be invariant under arbitrary automorphism, so if we have the statement I asked in the title, we get $G_A^{00}=G_{\varnothing}^{00}$.
I solved my own question in the following way, which did not give a full answer to the question in the title, but already helps solve the question about $G_A=G_{\varnothing}^{00}$:
The statement is correct for sets defined by complete types, but may not be true for sets defined by partial types (I am lacking in a counter-example).
Let $\Phi(x)$ be a partial type invariant under any automorphism. Let $p$ be any completion of $\Phi(x)$. For any $a$ realizing $p$ and $a'\models p|_{\varnothing}$, $a$ and $a'$ are conjugate over $\varnothing$. Moreover, $a\models\Phi(x)$, so by invariance of $\Phi(x)$ under any automorphism, $a'\models\Phi(x)$. This means that $p|_{\varnothing}\vdash\Phi(x)$. If $\Phi(x)$ is complete, we will have $\Phi(x)\vdash p|_{\varnothing}$, but for a strict partial type $\Phi(x)$, we may not have that.
