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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?

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    @Wox : If you require automorphisms to preserve an inner product, it means you're looking at the lattice as $\mathbb{Z}$-module **and** a metric space. From that point of view, square and rectangular lattices are **not** the same (they are not isometric). So it's not a big surprise that their automorphism group turn out to be different. Remember that the expression of the inner product changes with the basis. So choosing the "right" basis does indeed make the lattice isomorphic to $\mathbb{Z}^n$ as a $\mathbb{Z}$-module, but messes up the inner product doing so.2011-10-10

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That you feel there should be a difference between square/cubic and other lattices may indicate that you're not making a sufficiently clear distinction between the linear structure and the norm / metric / inner product space structure. The square lattice differs e.g. from the hexagonal lattice in the number of nearest neighbours of a point, and thus in the order of isometries around a point, but the notions of "nearest neighbour" and "isometry" are not linearly invariant. You can apply a linear skew transformation to the square lattice to turn it into a hexagonal lattice, changing its metric structure but leaving its linear structure invariant.