Let $k$ be a field. Let $\bar k$ be an algebraic closure of $k$. Let $G = \mbox{Aut}(\bar k/k)$. Let $n \ge 1$ be an integer. $G$ acts on $\bar k^n$ in the obvious way. Let $V$ be an irreducible algebraic set in $\bar k^n$. Let $\sigma \in G$. It is easy to see that $\sigma(V) = \{\sigma(x)\colon x \in V\}$ is an irreducible algebraic set. Hence $G$ acts on the set of irreducible algebraic sets in $\bar k^n$.
Is the following proposition true? If yes, how do we prove it?
Proposition. Let $\mathfrak p$ be a prime ideal of the polynomial ring $k[X_1, \dots, X_n]$. Let $V$ be the algebraic set in $\bar k^n$ defined by $\mathfrak p$. Let $V_1, \dots, V_r$ be the irreducible components of $V$. Then $G$ acts transitively on the set $\{V_1, \dots, V_r\}$. Moreover, $\dim V_i = \dim k[X_1, \dots, X_n]/\mathfrak p$ for all $i$.
Conversely let $W$ be an irreducible algebraic set in $\bar k^n$. Then the $G$-orbit $\{\sigma(W)\colon \sigma \in G\}$ is finite. Let $V = \bigcup_{\sigma\in G} \sigma(W)$. Then there exists a prime ideal $\mathfrak p$ of $k[X_1, \dots, X_n]$ such that $V$ is the algebraic set defined by $\mathfrak p$.