Why can't you simulate isotropic fluid flow on a square lattice?

It seems that this might be the original source: Lattice gas automata for the Navier-Stokes equations, by Frisch et al. (PDF)

The [square lattice] HPP automaton is invariant under $\pi/2$ rotations. Such a lattice symmetry is insufficient to ensure the isotropy of the fourth degree tensor relating momentum flux to quadratic terms in the velocity. …
Observe that when the underlying microworld is two-dimensional and invariant under the hexagonal rotation group (multiples of $\pi/3$) the tensor $T$ is isotropic…

Unfortunately it will be a long time before I am able to understand this properly; I would still prefer to see an answer that explains what is going on.

As you say you are not very familiar with tensors (and I wouldn't call myself an expert either), I will try to explain everything in linear-algebraic language. To quote the paper,

...for the HPP model, the momentum flux tensor has the form $$P_{\alpha\beta} = p\delta_{ab} + T_{\alpha\beta\gamma\epsilon}u_\gamma u_\epsilon + O(u^4).$$ ...The tensor $T$ is, by construction, symmetric in both $(\alpha,\beta)$ and $(\gamma,\epsilon)$.

Here $T$ is "the fourth degree tensor relating momentum flux to quadratic terms in the velocity". We can think of the second-order tensors $P_{\alpha\beta}$ and $u_\gamma u_\epsilon$ as symmetric $2\times2$ matrices ($P_{\alpha\beta}$ must be symmetric to conserve angular momentum, while $u_\gamma u_\epsilon$ is $uu^T$ which is clearly symmetric), and $T$ as a linear transformation mapping the latter to the former.

A tensor is isotropic if it looks the same under any rotation of the coordinate frame. In this case, that means that if $T$ maps some matrix $X$ to a matrix $Y$, and you rotate the coordinate frame so that $X$ becomes $X'$ and $Y$ becomes $Y'$, then $T$ still maps $X'$ to $Y'$. Formally, if a rotation transforms vectors as $v' = Rv$, then matrices transform as $A' = RAR^T$, and we want $T(RXR^T)=RT(X)R^T$.

Things become clearer if we make a judicious choice of basis for the three-dimensional space of symmetric $2\times2$ matrices, namely $$E_1 = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix},\quad E_2 = \begin{bmatrix}1 & 0\\0 & -1\end{bmatrix},\quad E_3 = \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}.$$ Then it turns out that a rotation by an angle $\theta$ transforms this basis as $$\begin{align} E_1' &= E_1, \\ E_2' &= E_2\cos2\theta + E_3\sin2\theta, \\ E_3' &= E_3\cos2\theta - E_2\sin2\theta, \end{align}$$ which one can also see by visualizing them as quadratic functions of the form $x \mapsto x^TEx$ over $\mathbb R^2$. In other words, a rotation transforms $E_2$ and $E_3$ into each other, so for $T$ to be invariant under such a transformation, it has to do "the same thing" to both $E_2$ and $E_3$, in some sense. But if you only look at rotations of $\pi/2$, you don't see this dependence, because then $E_2' = -E_2$ and $E_3' = -E_3$. So a tensor that treated $E_2$ and $E_3$ differently could be invariant under rotations of $\pi/2$, but it would not be invariant under arbitrary rotations. For example, we could choose $T$ such that $$T(E_1) = T(E_2) = 0,\quad T(E_3) = E_3.$$ You can verify that this tensor is invariant under a rotation of $\pi/2$, but not under a rotation of $\pi/4$ (where $E_2' = E_3$ and $E_3' = -E_2$), so it is not isotropic.

Anyway, since the title of the question still asks "Why can't you simulate isotropic fluid flow on a square lattice?", allow me to reiterate my comment that this is an issue with lattice gas automata in particular, not with fluid simulations on square grids in general. The problem with isotropy comes in when trying to show the convergence of the cellular automaton model to the Navier-Stokes equations. As far as I am aware, you can simulate isotropic fluid flow on a square grid through the usual methods of directly discretizing the Navier-Stokes equations and solving them numerically, without going through an intermediate cellular automata model.