A characterization of the subgroup of $\text{GL}(\bigwedge^k V)$ which preserves pure tensors? Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$. Set
$$H=\{B\in\text{GL}(\bigwedge^k V)   \,   |   \, B \,\text{  preserves pure tensors }\}$$
(i.e. $B \in H$ if it maps decomposable elements to decomposable elements).

$H$ is a closed subgroup of the Lie group $\text{GL}(\bigwedge^k V) $, hence it is an embedded Lie subgroup of it. 
  Can we characterize it explicitly? What is $\dim H$? How does its Lie algebra look like?

Here is a lower bound: The exterior power map $\psi:\text{GL}( V) \to \text{GL}(\bigwedge^k V)$ defined by $\psi(A)=\bigwedge^k A$ is a smooth immersion*. Since $\text{Image}(\psi)\subseteq H$, $\dim H \ge  \dim \text{GL}( V) =d^2$.

I think that this might be related to the following question: Let $\omega^I$ be a basis for $\bigwedge^kV$, whose elements are all decomposable. Is $\omega^I$ "induced" by a standard basis, i.e. $\omega^{i_1,\ldots,i_k}=v^{i_1} \wedge \ldots \wedge v^{i_k}$ for some basis $v_i$?

*One way to see that the exterior power map $\psi$ is an immersion, is that it is a locally-injective homomorphism of Lie groups, and in particular, it has a constant rank. The local injectivity (which comes from the fact that $\bigwedge^k A=\bigwedge^k B \Rightarrow A= \pm B$) then implies that $\psi$ is an immersion, due to the constant rank theorem.
 A: 
Here is a solution for the case where $V$ is a complex vector space. (I guess that the real case can be reduced to the complex one, see some subtle points below). It turns out that $H$ is pretty restricted: When $d \neq 2k$, $H=\text{Image}(\psi)$, i.e. every map which preserves pure tensors is an exterior power of some $A \in \text{GL}$.
When $d=2k$, we have additional automorphisms, of the form $\star \circ \bigwedge^k A$, where $\star:\bigwedge^k V \to \bigwedge^k V$ is the Hodge dual operator w.r.t some fixed metric on $V$.

For a reference, see here.
Here are some natural questions which  arise from this representation claim:
1. Is this representation redundant in some way?
i.e. if $\star$ equals to some exterior power, then it is not a "true" addition to the subgroup $H$. However, the Hodge is never an exterior power: Indeed, suppose that $\star=\bigwedge^k A$ for some $A \in \text{GL}(V)$. Let $v_1,\dots,v_d$ be an orthonormal basis for $V$.
$\text{span}(Av_1,\dots,Av_k)=\text{span}(v_{k+1},\dots,v_d)$. Switching $v_k$ and $v_{k+1}$, we also deduce that $\text{span}(Av_1,\dots,Av_{k-1},Av_{k+1})=\text{span}(v_k,v_{k+2},\dots,v_d)$, so $$\text{span}(Av_1,\dots,Av_{k-1})= \text{span}(v_{k+2},\dots,v_d).$$
In particular, we have proved that for any orthonormal basis $v_1,\dots,v_{k-1},v_{k},v_{k+1},\dots,v_d$, $$\text{span}(Av_1,\dots,Av_{k-1}) \subseteq \text{span}(v_{k+2},\dots,v_d). \tag{1} $$ Now, switching between $v_{k+1}$ and $v_{k+2}$, this also implies
$$\text{span}(Av_1,\dots,Av_{k-1}) \subseteq \text{span}(v_{k+1},v_{k+3},v_{k+4},\dots,v_d) \tag{2}.$$
Combining $(1)$ and $(2)$ together, we deduce that
$$\text{span}(Av_1,\dots,Av_{k-1}) \subseteq \text{span}(v_{k+3},\dots,v_d) $$, which implies that $A$ maps the $k-1$ dimensional space $\text{span}(v_1,\dots,v_{k-1})$ to the $k-2$ dimensional space $\text{span}(v_{k+3},\dots,v_d) $, i.e. $A$ is not invertible. This implies $\bigwedge^k A$ is also singular, so it cannot be equal to a Hodge dual. Contradiction.
2. What about automorphism of the form $ \bigwedge^k A \circ 
 \star $? or $ \star \bigwedge^k A \circ 
 \star $?
Well, using the equality $\star \circ \bigwedge^k A=\det A \bigwedge^k A^{-T} \circ \star=\bigwedge^k (\sqrt[k]{\det A}\cdot A^{-T} )\circ \star$ we can "move" $\star$ from one side of $\bigwedge^k A$ to the other one. (Here we used the assumption that $V$ is a complex vector space, since otherwise $\sqrt[k]{\det A}$ does not exist if $k$ is even, and $\det A<0$). Maybe in that case we need to allow multiplication of exterior powers by negative scalars).
3. How do we account for all the different Hodge duals of all the  different metrics, by using the Hodge operator induced by a single metric?
Since the map $g \to \star_g$ is smooth and injective up to a conformal rescaling of the metric (that is $\star_g=\star_h$ if and only if $g=ch$ ), one expects to obtains a continuous family of different Hodge duals of dimension $\dim(\text{Psym_d})-1$, so the dimension of $H$ should increase accordingly. However, this is not really what happens, since we can or express every Hodge dual in terms of one specific representative, by compensate for the difference via some exterior power of a mediating map.
(Actually, I think that we can induce a Hodge dual using any non-degenerate bilinear form, symmetric or not). Here are the details:
Suppose $g,h$ are two metrics on $V$. Then we can write $g(v,w)=h(v,Tw)$ for some $T \in \text{GL}(V)$. It is obvious that $T$ is $h$-positive, and $h$-symmetric, hence there exist a uniqe $\sqrt T$ which is also $h$-positive and symmetric. Thus, we have $g(v,w)=h(\sqrt Tv,\sqrt Tw)$, i.e. $\sqrt T:(V,g) \to (V,h)$ is an orientation-preserving isometry, hence it commutes with the Hodge dual, i.e.
$\bigwedge^k \sqrt T \circ \star_g= \star_h \circ \bigwedge^k \sqrt T$. So, now we can write
$$ \star_g \circ \bigwedge^k A=\bigwedge^k \sqrt T^{-1} \circ \star_h \circ \bigwedge^k \sqrt T A,$$
and since we only know how to move $\star_h$ from one side of an exterior power to the other side (see point $2$), this representation can be reduced to $\star_h \circ \bigwedge^B$ for some $B \in \text{GL}(V)$.
Similarly, we can represent creatures of the form of $\star_h \circ \star_g$...by expressing $\star_g$ via $\star_h$ or vice-versa.
