Concider a lattice in $\mathbb R^{n}$ (i.e. all $\mathbb Z$-linear combinations of a chosen basis). The coordinates of all vectors in this lattice with respect to its basis, is given by $\mathbb Z^{n}$. Am I right in stating that the automorphism group of a lattice consists of all permutations ($n!$) and sign changes ($2^n$) of $\mathbb Z^{n}$? This would mean that the order of the automorphism group is $2^nn!$, regardless of which lattice I'm talking about (e.g. in $\mathbb R^{2}$ you have the square, hexagonal, rectagonal, ... lattice).
I suspect this only goes for the square lattice (or cubic lattice in $\mathbb R^{3}$), but I can't understand why. Choosing the right basis for any lattice makes it isomorphic to $\mathbb Z^{n}$ right?