Let the group $G$ act on the set $X$.
For $g \in G$ and $A \in P(X)$ set $g.A = \{g.a \mid a \in A\} = \{x \in X \mid \exists a \in A: x = g.a \}$. this defines an action of $G$ on $P(x)$. I 'm trying to show that this action is an automorphism given the map $\phi: A \mapsto g.A$, where $\phi: (P(x),\Delta) \rightarrow (P(x),\Delta)$.
Here, $A \Delta B = (A - B) \cup (B - A)$ the symmetric difference of $A$ and $B$.
From the definitions, I know I have to show that for $A,B \in P(X)$, we have to show that $\phi (A \Delta B) = \phi(A) \ \Delta \ \phi(B)$. However, I'm unsure where to go from here.