= Solution
With constant $\nu$, the bulk <action> is invariant under time translations, translations in the two spatial coordinates, and rotations or reflections within the substrate plane. The <Laplacian> has no preferred direction, giving the planar $O(2)$ symmetry. It also has <scalar-field shift symmetry> $h\mapsto h+a$, expressing the arbitrary origin of height, and the discrete symmetry $h\mapsto-h$ because its residual is squared.
There is an additional exact tilt symmetry $h\mapsto h+\mathbf b\cdot\mathbf x$: the added slope is time independent and has zero <Laplacian>. More generally, adding any function $f$ satisfying $f_t=\nu\nabla^2f$ leaves the residual unchanged. Such bulk transformations need not preserve prescribed boundary conditions or a restricted physical interpretation of the height.
In two spatial dimensions the <action> also has the anisotropic <scale invariance>
$$
\boxed{\mathbf x\mapsto b\mathbf x,\qquad t\mapsto b^2t,\qquad h\mapsto h\quad(b>0).}
$$
The spacetime measure scales as $b^4$ and the squared residual as $b^{-4}$. The time and length scales are therefore related diffusively, rather than by a Lorentz transformation.
Ordinary <time reversal> needs a boundary qualification. The identity
$$
(h_t-\nu\nabla^2h)^2=h_t^2+\nu^2(\nabla^2h)^2+\nu\partial_t|\nabla h|^2-2\nu\nabla\cdot(h_t\nabla h)
$$
shows that the density is time-reversal invariant up to a <total divergence>. Thus the bulk stationary equation $h_{tt}-\nu^2\nabla^4h=0$ is invariant under $t\mapsto-t$, although the printed density is not pointwise unchanged. The <time reversal of the squared diffusion-residual action> exchanges the growing and decaying branches; the selected first-order <diffusion equation> alone is not time-reversal invariant at fixed positive $\nu$.
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