@MarcvanLeeuwen I'm a young man.I was so eager to solve a problem and needed this lemma.I didn't realize it is unmoral to ask the queation simutaneously on MO and math.stackexchange.I feel ashamed now and will be cautious.Please forgive me.

@GerryMyerson I'm sorry,Gerry Myerson.I wll be cautious from now on.I also appologize to everyone for my levity.

no,say $n=1$,$Aut(X)$ should be the cyclic group generated by $i$.But according to your answer,$Aut(X)=\mathbb{Z}[i]\setminus \{0\}$,so you are wrong.

I don't know why $F$ has to be of that form,can you explain it at length?Moreover,since $F\in Aut(G)$,which is a group,I even can't check how to determine the coefficients to let $F$ be the identity.So I think there must be something wrong assuming that the discrete subgroup is given by the matrices whose entries z1, z2, z3 are Gaussian integers.