A map which commutes with Hodge dual is conformal? $\newcommand{\Hom}{\operatorname{Hom}}$
$\newcommand{\Cof}{\operatorname{Cof}}$
$\newcommand{\Det}{\operatorname{Det}}$
$\newcommand{\id}{\operatorname{Id}}$
Let $V$ and $W$ be $d$-dimensional, oriented inner-product spaces, and let $A\in\Hom(V,W)$ be an orientation-preserving map. Suppose that $A$ satisfies 
$$
\bigwedge^{d-k} A  \circ \star_V^k= \star_W^{k} \circ \bigwedge^k A \neq 0, \tag{1}
$$
for a single $1 \le k \le d-1$.

Question: Is it true that $A$ is conformal? If not, can we characterise the maps which satisfy this?

Comment (1): Condition $(1)$ implies $A$ is invertible.  (Proof at the end). If we omit the part $\bigwedge^k A \neq 0$, then every map with rank less than $\min(k,d-k)$ would satisfy the commutation property (and won't be conformal).
Comment (2): Condition $(1)$ is symmtric in $k,d-k$.
Comment (3):  I guess that if $k \neq d-k$, then $A$ needs to be an isometry. (The heuristics is this: Condition $(1)$ says the action on $A$ on parallelepipeds of dimensions $k,d-k$ is "the same", and when the scales are different, this is a rigidity constraint).
Edit(1): I proved that if $A$ is conformal, and $k \neq d-k$, then $A$ is an isometry, so we are back at the conformality question.
Partial progress (reformulation and proof for $k=1,d-1$):
Define $\Cof^k A := (-1)^{k(d-k)} \star_W^{d-k} \circ \bigwedge^{d-k} A \circ \star_V^k$. Note that $\Cof^k A \in\Hom( \Lambda_k(V) ,\Lambda_k(W))$. It can be proved that
$$\Det A \cdot \id_{\Lambda_k(V)} = \bigwedge^k A^T \circ \Cof^k A \tag{2}$$
for any map $A \in \Hom(V,W)$. 
Condition $(1)$ implies that $\Cof^k A = \bigwedge^k A$. Plugging this into equation $(2)$, we get
$$ \Det A \cdot \id_{\Lambda_k(V)} = \bigwedge^k A^T \circ \bigwedge^k A =\bigwedge^k (A^T \circ A) \tag{3}.$$
Denote $A^TA=S \in \Hom(V,V)$. Then we obtained 

$$ \bigwedge^k (\Det S)^{\frac{1}{2k}}\id_V  =\sqrt{\Det S} \cdot \id_{\Lambda_k(V)}  =\bigwedge^k S \tag{4}.$$
Does this imply $S=(\Det S)^{\frac{1}{2k}}\cdot \id_V$?

If the wedge was "injective" in the sense that $\bigwedge^k S=\bigwedge^k T \Rightarrow S=T,$ we were done. Of course, this injectivity does not hold in general, since for $S=0$ any map $T$ of rank smaller than $k$ would satisfy $\bigwedge^k T=0$.
For $k=1$ the injectivity holds, so we are done. (We get $A^TA=\det A \id_V$. If $\det A= 0$ we get $A^TA=0 \Rightarrow A=0$ which contradicts the assumption. So  $\det A \neq 0$, and $A$ is conformal).
Since condition $(1)$ is symmetric in $k,d-k$, the answer is also positive for $k=d-1$ 

Proof that condition $(1)$ implies invertibility:
Suppose by contradiction $\ker A \neq 0$, and let $r=\operatorname{rank}(A)=\dim\big((\ker A)^{\perp}\big)$. Since $ \bigwedge^k A \neq 0 \Rightarrow \operatorname{rank}(A) \ge k$, condition $(1)$ implies $r \ge k$. Let $v_1,\dots,v_r$ be an orthonormal basis for $(\ker A)^{\perp}$, and let $w_1,\dots,w_{d-r}$ be an orthonormal basis for $\ker A$.
Then
$$\bigwedge^{d-k} A  \circ \star_V^k (v_1 \wedge \dots \wedge v_k)= \star_W^{k} \circ \bigwedge^k A (v_1 \wedge \dots \wedge v_k)$$
Since $A|_{(\ker A)^{\perp}}$ is injective, the RHS is non-zero, while
the LHS equals $\pm \bigwedge^{d-k} A (v_{k+1} \wedge \dots \wedge v_r \wedge w_1 \dots \wedge w_{d-r})=0 $, a contradiction.
 A: Here is a proof for "diagonal maps" $A:V \to V$:
(The general case can probably be reduced to this case using SVD):
Let $v_1,...,v_d$ be a positive orthonormal basis for $V$, and suppose $Av_i=\sigma_iv_i, \sigma_i \ge 0$. By the remark in the question, we can assume $A$ is invertible so all $\sigma_i > 0$.
Then
$$\bigwedge^{d-k} A  \circ \star_V^k (v_{i_1} \wedge \dots \wedge v_{i_k})=\bigwedge^{d-k} A  (v_{i_{k+1}} \wedge \dots \wedge v_{i_d})= Av_{i_{k+1}} \wedge \dots \wedge Av_{i_d}=$$
$$\sigma_{i_{k+1}} v_{i_{k+1}} \wedge \dots \wedge \sigma_{i_d} v_{i_d}=\Pi_{j=k+1}^d \sigma_{i_j} \cdot (v_{i_{k+1}} \wedge \dots \wedge v_{i_d}).$$
On the other hand,
$$ \star_V^{k} \circ \bigwedge^k A (v_{i_1} \wedge \dots \wedge v_{i_k})=\star_V^{k} \big( \Pi_{j=1}^k \sigma_{i_j} \cdot (v_{i_1} \wedge \dots \wedge v_{i_k}) \big)= \Pi_{j=1}^k \sigma_{i_j}(v_{i_{k+1}} \wedge \dots \wedge v_{i_d}).$$
Now, the condition  $\bigwedge^{d-k} A  \circ \star_V^k= \star_V^{k} \circ \bigwedge^k A $ implies $$\Pi_{j=1}^k \sigma_{i_j}=\Pi_{j=k+1}^d \sigma_{i_j}.$$
for any choice of indices $1 \le i_1,\dots,i_k \le d$. Write explicitly $$ \sigma_{i_1} \cdot \sigma_{i_2} \cdot \dots \cdot \sigma_{i_k}=\sigma_{i_{k+1}} \cdot \sigma_{i_{k+2}} \cdot \dots \cdot \sigma_{i_d}, \tag{1}$$ and switch $i_k$ and $i_{k+1}$, so 
$$ \sigma_{i_1} \cdot \sigma_{i_2} \cdot \dots \cdot \sigma_{i_{k+1}}=\sigma_{i_k} \cdot \sigma_{i_{k+2}} \cdot \dots \cdot \sigma_{i_d} \tag{2}$$ also holds.
Dividing $(2)$ by $(1)$, we get 
$$ \frac{\sigma_{i_k}}{\sigma_{i_{k+1}}}=\frac{\sigma_{i_{k+1}}}{\sigma_{i_{k}}} \Rightarrow \sigma_{i_k}=\sigma_{i_{k+1}}.$$
Since $i_k,i_{k+1}$ were arbitrary indices, we conclude all the $\sigma_i$ are equals, so $A$ is conformal.
Note: The proof indeed "breaks" when $k=0,d$ since we cannot switch two indices between different sides of the equality. In that case the commutation property is equivalent to $\det A=1$.
A: $\newcommand{\Hom}{\operatorname{Hom}}\newcommand{\Cof}{\operatorname{Cof}}\newcommand{\Det}{\operatorname{Det}}\newcommand{\id}{\operatorname{Id}}$Here is an "alternative" answer, which is a bit simpler than the original:
Instead of working (implicitly) with SVD, we note the main claim was reduced (in the question) to the following:
Let $S \in \Hom(V,V)$ be symmetric and positive-definite (i.e $S^T=S$, $S$ has positive eigenvalues). Suppose 

$$ \bigwedge^k \id_V  = \id_{\Lambda_k(V)}  =\bigwedge^k S \tag{4}.$$
Then $S=\id_V$.


Comment: This claim does not hold if we relax the positivity requirement, i.e take $S=-\text{Id}$, $k$ even.
Proof:
By orthogonal diagonalization,we can assume $S$ is diagonal:
$sv_i=\sigma_i v_i,\sigma_i > 0$, where $v_1,...,v_d$ is an orthonormal basis for $V$.
$$v_{i_1} \wedge \dots \wedge v_{i_k}=\bigwedge^{k} S (v_{i_1} \wedge \dots \wedge v_{i_k})=(\Pi_{j=1}^k  \sigma_{i_j}) (v_{i_1} \wedge \dots \wedge v_{i_k}),$$
implies $ \Pi_{j=1}^k  \sigma_{i_j}=1,$ for any choice of indices $1 \le i_1,\dots,i_k \le d$. Write explicitly $$ \sigma_{i_1} \cdot \sigma_{i_2} \cdot \dots \cdot \sigma_{i_{k-1}} \cdot  \sigma_{i_k}=\sigma_{i_1} \cdot \sigma_{i_2} \cdot \dots \cdot \sigma_{i_{k-1}} \cdot \sigma_{i_{k+1}}$$ 
(we switched $i_k$ and $i_{k+1}$), we deduce $\sigma_{i_k}=\sigma_{i_{k+1}}.$
Since $i_k,i_{k+1}$ were arbitrary indices, we conclude all the $\sigma_i=\sigma$ are equals. Now $\sigma^k=\Pi_{j=1}^k  \sigma_{i_j}=1$ implies $\sigma=1$, so $S=\id_V$.
