# Automorphisms of a lattice and changing to a nicer $\mathbb{Z}$-base

Suppose I have an integral lattice $L$ with an arbitrary $\mathbb{Z}$-base, equipped with a positive-definite nondegenerate symmetric bilinear form $\langle\cdot,\cdot\rangle$, and an isometry $\nu$ on $L$ that preserves $\langle\cdot,\cdot \rangle$. My question is this: Is there necessary and sufficient conditions that can guarantee/allow me find a "nicer" $\mathbb{Z}$-base, where $\nu$ acts on this base by either permuting base elements, or sending $\alpha\rightarrow -\alpha$, or fixes the base element, or a composition thereof.

I don't think that I am saying this right so I let me give an example. Let's say $L$ is rank $5$ and I have knowledge that $\nu$ is of order 3 on one element, fixes 2 elements, and acts as $-1$ on another element. Is there way I know whether there exists elements $\alpha_1, \nu\alpha_1=\alpha_2$, $\nu^3\alpha_1=\alpha_1$, $\nu\alpha_4= -\alpha_4$, and $\nu\alpha_5=\alpha_5$, such that $\{\alpha_i\}$ spans $L$ and is not just a sublattice. What if I take the nondegenerate condition out? I ask this because I am having a hard time finding both a proof or counter example to this.

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