Prove that $\displaystyle Aut (Z_{2}XZ_{2}) \cong S_{3} $.
Let $\displaystyle \phi: \mathbb{Z}_2\times \mathbb{Z}_2\to \mathbb{Z}_2\times \mathbb{Z}_2$ be an automorphism.
Obviously, $\displaystyle \phi(0,0) = (0,0)$.
1)Now, $\displaystyle \phi(0,1)$ has to be an element of order $\displaystyle 2$ and so $\displaystyle \phi(0,1) = (0,1) \text{ or }(1,0)\text{ or }(1,1)$.
2)The same possibilities for $\displaystyle \phi(0,1)$ however except for what was used in #1.
3)Once #1 and #2 are determined then $\displaystyle \phi(1,1) = \phi(1,0) + \phi(0,1)$ and so $\displaystyle \phi(1,1)$ is determined.
There are $\displaystyle 3$ possibilities for #2 and $\displaystyle 2$ possibilities #3, we therefore have at most $\displaystyle 2\cdot 3 = 6$ automorphisms.
Check that all these six instead give a raise to a hextic automorphism group.