Affine algebraic varieties defined over a [not necessarily algebraically closed] field 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 $\mathfrak p$.
 A: After posting some nonconstructive comments I decided to post a constructive answer.
Set $A=k[X_1, \dots, X_n]$ and $B=\bar k[X_1, \dots, X_n]$. The ring extension $A\subset B$ is flat (why?). Let $\mathfrak p$ be a prime ideal of $A$. The irreducible components of $V$, the algebraic set in $\bar k^n$ defined by $\mathfrak p$, are defined by the  prime ideals $P_1,\dots,P_r$ of $B$ which are minimal over $\mathfrak pB$. Since the extension $A\subset B$ is flat it follows that $P_i\cap A=\mathfrak p$ for all $i$. Moreover, the ring extension $A_{\mathfrak p}\subset B_{P_i}$ is also flat and now we can apply the dimension formula and get $\dim B_{P_i}=\dim A_{\mathfrak p}+\dim B_{P_i}/\mathfrak pB_{P_i}$. But $\dim B_{P_i}/\mathfrak pB_{P_i}=0$ (why?) and thus we get $\dim B_{P_i}=\dim A_{\mathfrak p}$, that is, $\mbox{ht}(P_i)=\mbox{ht}(\mathfrak p)$ and this is enough to show that $\dim V_i = \dim k[X_1, \dots, X_n]/\mathfrak p$. 
The extension $A\subset B$ is an integral extension of integrally closed domains. Furthermore, the field extension $K\subset L$ is normal, where $K$ and $L$ are the fields of fractions of $A$, respectively $B$, so we can apply Theorem 5(vi), page 33, from Matsumura, CA, that says the following: any two prime ideals $P_1$ and $P_2$ of $B$ lying over the same prime ideal $\mathfrak p$ of $A$ are conjugate, i.e. there is $\bar\sigma\in G(L/K)$ such that $\bar\sigma(P_2)=P_1$. Now take $\sigma=\bar\sigma_{\mid \bar k}\in G$. If I'm not wrong, $\sigma(V_2)=V_1$.
For the converse, let $P$ be a prime ideal of $B$ that defines $W$ and $\mathfrak p=P\cap A$. It remains to prove that $W$ is the algebraic set of $\bar k^n$ defined by $\mathfrak p$. This can be reformulated as follow: the prime ideals of $B$ lying over $\mathfrak p$ are the minimal elements of the set of the prime ideals of $B$ containing $\mathfrak p$, and this is easy to prove by using the well-known properties of integral extensions. 
