Let $V$ be a finite-dimensional real inner product space and let $L \subset V$ be a lattice of full rank. Consider the finite group
$$ \mathrm{Aut}(L) = \{ f \in \mathrm{O}(V) \mid f(L) = L \}, $$
which we call the group of automorphisms of $L$.
Question. When does $\mathrm{Aut}(L)$ act irreducibly on $V$? Can we formulate a necessary and/or sufficient condition for that purely in terms of the lattice $L$ (say, in terms of the Gram matrix of some/any basis of it)?
Intuitively, we want the lattice to be sufficiently symmetric. For example, one can show that $\mathbb{Z}^n \subset \mathbb{R}^n$ does the trick. It is also easy to show that if $R \subset V$ is an irreducible root system, then the Weyl group $W(R) \subseteq O(V)$ acts irreducibly on $V$ and thus $\mathrm{Aut}(L) \supset W(R)$ acts irreducibly on $V$ as well, where $L \subset V$ is the lattice generated by $R$. We can even allow the root system to be mildly reducible: if $R \subset V$ is an irreducible root system, consider the root system $R^k = \underbrace{R \sqcup \ldots \sqcup R}_\text{$k$ times} \subset \underbrace{V \oplus \ldots \oplus V}_\text{$k$ times} = V^k$, where the inner product on the latter is just $k$ copies of the inner product on $V$ and the direct sum decomposition is orthogonal. Then the group $\mathrm{Aut}(R^k)$ of automorphisms of $R^k$ as a root system acts irreducibly on $V^k$ and thus $\mathrm{Aut}(L) \supset \mathrm{Aut}(R^k)$ acts irreducibly on $V^k$, where $L \subset V^k$ is the lattice generated by $R^k$ (this examples includes $\mathbb{Z}^n \subset \mathbb{R}^n$ as the lattice generated by the root system $A_1^n$).
On the other hand, one can easily come up with an example of a lattice whose automorphism group acts reducibly on the ambient space. For instance, consider a lattice in $\mathbb{R}^2$ spanned by the vectors $(1,0)$ and $(0,2)$. Its automorphism group is just $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ generated by the reflections in the coordinate axes, and it acts reducibly on $\mathbb{R}^2$.
Motivation. I arrived at this question by looking at flat tori and their isotropy groups and trying to understand which flat tori are isotropy-irreducible. Each flat torus is of the form $T = \mathbb{R}^n/L$, where $L$ is a lattice of full rank, and it is easy to see by looking at the universal Riemannian covering $\mathbb{R}^n \twoheadrightarrow T$ that the isometry group of $T$ is isomorphic to $\mathrm{Aut}(L) \ltimes T$. Consequently, the isotropy subgroup of $Isom(T)$ at the base point $0 + L$ is isomorphic to $\mathrm{Aut}(L)$, so $T$ is isotropy-irreducible precisely when the automorphism group of $L$ acts irreducibly on $\mathbb{R}^n$.