Let $V$ be a real oriented $d$-dimensional inner product space, $d \ge 3$. For $1 \le k \le d-1$, the Hodge dual map $\star: \bigwedge^k V \to \bigwedge^{d-k} V$ commutes with orientation-preserving isometries:
For every $Q \in \text{SO}(V)$, we have $$\star \circ \bigwedge^k Q= \bigwedge^{d-k} Q \circ \star \tag{1}.$$
Is $\star$ the unique linear map $\bigwedge^k V \to \bigwedge^{d-k} V$ satisfying $(1)$ up to scaling?
In the language of representation theory, I ask if the space of equivariant maps w.r.t the natural representations of $ \text{SO}(V)$ on $\bigwedge^k V,\bigwedge^{d-k} V$ is one dimensional.
In $d=2$, $\star:V \to V$ is of course not the unique map up to scaling which commutes with all isometries, since $\text{SO}(2)$ is commutative, we have additional elements... (This is why I restricted $d \ge 3$).
This is true unless $d$ is even and $k=d-k=\frac{d}2$. This follows directly from the representation theory description you give in the question using Schur's lemma. Unless $d$ is even and $k=\frac{d}2$ the representation $\Lambda^kV$ is irreducible and so the isomorphism to $\Lambda^{d-k}V$ is unique up to a scalar multiple.If $d$ is even and $k=\frac{d}2$, then $\Lambda^kV$ is the direct sum of two non-isomorphic irreducible representations (the two eigenspaces of $*$) and you can choose independently choose sclalar factors on the two components (so there is a two parameter family of homomorphisms).