Are there natural isomorphisms of Grassmannian and Stiefel manifolds with their "duals"?

Question

Let $V$ be a finite $n$-dimensional real vector space, let $V^*$ denote its dual, let $Gr(V,k)$ denote the set of all $k$-dimensional subspaces of $V$, $St(V,k)$ denote the set of all $k$-frames of $V$ (sets of $k$ linearly independent vectors in $V$), let $Gr(V^*, k)$ and $St(V^*,k)$ denote the corresponding concepts for $V^*$.

Given a linear transformation $T: V \to W$, denote by $Gr(T,k)$ to be the induced map $Gr(V,k) \to Gr(W,k)$ and denote by $St(T,k)$ the induced map $St(V,k) \to St(W,k)$.

Implicit Claim: $Gr(\cdot,k)$ and $St(\cdot,k)$ are functors from the categories of finite-dimensional vector space to some other category whose objects are Grassmannians or Stiefel manifolds respectively.

It is a well-known fact that there is no natural isomorphism from $V \to V^*$.

Question: However, are there natural isomorphisms: $$\begin{array}{rcl} Gr(V,k) & \cong & Gr(V^*, n-k) \\ St(V,n) & \cong & St(V^*,n) \end{array} \quad? $$

One can define an isomorphism for the first intersecting the kernels of all non-zero linear functionals (remembering also that we can identify $V \cong V^{**}$). The case $n=1$ is easiest to see (and the one for which I am most confident that such an isomorphism actually exists).

For the second isomorphism, given any basis, just take its dual basis. Does this lead to a natural isomorphism of the Stiefel manifolds? Or is the definition of dual basis just convenient, but not natural?

The answer to the question could fail to be positive for several reasons: (1) the implicit claim is false, either about the existence of the claimed target categories, or about the existence of the claimed functors even when the claimed target categories do exist, (2) one of the two described isomorphisms above does not exist/is not well-defined, (3) one of the two isomorphisms, despite existing, is nevertheless not natural.

Note: A pointer to a reference will suffice for an answer. In fact, I would prefer an answer that spared as many details as possible, because if the isomorphisms are natural, then it would be a good exercise for me to prove all of the above myself. So yes/no will even suffice.

Answer 1

Given a $k$-dimensional subspace $X$ of $V$, consider the subspace

$$X^{\ast} = \{ f \in V^{\ast} : f(x) = 0 \forall x \in X \}$$

of $V^{\ast}$. This subspace is $(n-k)$-dimensional and this defines an isomorphism

$$\text{Gr}_k(V) \cong \text{Gr}_{\dim V-k}(V^{\ast})$$

which is natural in all reasonable senses. Taking dual bases is at best natural in the sense that both bases of $V$ and bases of $V^{\ast}$ have natural $GL(V)$ actions and taking dual bases is equivariant with respect to these.

Edit: "All reasonable senses" means the following: both sides are functorial in $V$, regarded as taking values in the category of finite-dimensional vector spaces and injective linear maps. It should be more or less clear how to interpret $\text{Gr}_k(-)$ as a functor on this category. We can interpret $\text{Gr}_{\dim(-)-k}((-)^{\ast})$ as a functor as follows. If $T : V \to W$ is an injective map between finite-dimensional vector spaces, the dual map

$$T^{\ast} : W^{\ast} \to V^{\ast}$$

is surjective. This gives a well-defined pullback map

$$\text{Gr}_{\dim V-k}(V^{\ast}) \ni Y \mapsto (T^{\ast})^{-1}(Y) \in \text{Gr}_{\dim W-k}(W^{\ast})$$

and now it's true that the isomorphism above is natural in this stronger sense, as an isomorphism between two functors on finite-dimensional vector spaces and injective maps. (We can remove the finite-dimensionality hypothesis but then we need to redefine $\text{Gr}_{\dim V-k}(V^{\ast})$ to mean codimension-$k$ subspaces of $V^{\ast}$.)