This is a compilation of some questions that appear when dealing with very simple notions from linear algebra. It occurred to me while trying to answer this question with an example of a poset having exactly one initial and one final objects, which isn't isomorphic to its opposite. In the end, I realized I'm not sure how to prove that the example I had in mind is indeed an example. So I decided to post it as a separate question and link it to the original one. I tried to write it in the language of categories, but it's mostly about linear algebra and posets; hope it doesn't bother too much.
Of course, there are many easy examples of posets with the property described above, for instance many finite ones. I wanted, though, to give an example with a little bit more substance. In general, whenever one has a given 'thing' that admits 'subthings', it's possible to define a poset of subthings of the thing. Rarely, I believe, one can expect the posets obtained this way to be isomorphic to their opposites. Just imagine this possibility with respect to the spectrum of a ring.
Nevertheless, let $V$ be a finite-dimensional real vector space. We can define a poset ${\bf Gr}(V)$ whose objects are subspaces $W\subset V$ and whose morphisms are the inclusions $W_1\rightarrow W_2$ between subspaces of $V$. Obviously, $0\subset V$ and $V\subset V$ are the initial and the final objects of ${\bf Gr}(V)$, respectively.
It turns out that ${\bf Gr}(V)$ is isomorphic to ${\bf Gr}(V)^{\bf op}$. Indeed, let $\langle\phantom{v},\phantom{v}\rangle:V\otimes V\rightarrow \Bbb R$ be a positive-definite symmetric bilinear form on $V$. Then $\langle\phantom{v},\phantom{v}\rangle$ can be thought as an isomorphism $\varphi:V\rightarrow V^*$ such that $\varphi^*=\varphi$.
1. Being an isomorphism $\varphi:V\rightarrow V^*$ such that $\varphi^*=\varphi$ simply means that the bilinear form is non-degenerate and symmetric, respectively; positive-definiteness is not encoded in this description. How can the signature of the form be expressed in these terms? Of course, one can fix an isomorphism $\psi:V\rightarrow V^*$, bring $\varphi$ to an automorphism of $V$ by means of $\psi$, translate the action on forms to mere conjugation, and look at the invariants of this action. But, how is it possible to get rid of the choice of $\psi$ ?
By abuse of notation, we also denote by $\varphi:W\rightarrow W^*$ the isomorphism induced by $\varphi:V\rightarrow V^*$, where $W\subset V$ is a subspace. (Here's where positive-definiteness plays a role. Exercise: why? Hint: what is ${\rm Ker}(m^*\circ\varphi)\subset V$ in geometrical terms?)
$$\require{AMScd} \begin{CD} W@>m>>V\\ @V\varphi VV@V\varphi VV\\ W^*@<m^*<<V^* \end{CD} $$
It's easy to see that the functor ${\bf Gr}(V)\rightarrow{\bf Gr}(V)^{\bf op}$ defined by the formula $m\mapsto\varphi^{-1}m^*\varphi$ is an isomorphism.
$$\require{AMScd} \begin{CD} W_1@>m>>W_2\\ @A\varphi^{-1}AA&@V\varphi VV\\ W_1^*@<m^*<<W_2^* \end{CD} $$
2. Is it possible to (slightly or radically) modify the definition of ${\bf Gr}(V)$ so that ${\bf Gr}(-)$ becomes a functor (or some related notion) on the category of vector spaces (or some related category)? If it is, then is ${\bf Gr}(-)$ naturally isomorphic to ${\bf Gr}(-)^{\bf op}$ ? If it isn't, then why not?
Now, let ${\rm GL}(V)$ be the group of linear automorphisms of the vector space $V$. We can define a poset ${\bf GL}(V)$ whose objects are subgroups $G\subset {\rm GL}(V)$ and whose morphisms are the inclusions $G_1\rightarrow G_2$ between subgroups of ${\rm GL}(V)$. Obviously, $1\subset{\rm GL}(V)$ and ${\rm GL}(V)\subset{\rm GL}(V)$ are the initial and the final objects of ${\bf GL}(V)$, respectively.
Clearly, there's some relation between ${\bf Gr}(V)$ and ${\bf GL}(V)$. Indeed, to a subspace $W\subset V$ we can associate a subgroup ${\rm Stab}_{{\rm GL}(V)}(W)\subset{\rm GL}(V)$ consisting of the elements of ${\rm GL}(V)$ that stabilize $W$. Of course, this rule doesn't define an isomorphism between ${\bf Gr}(V)$ and ${\bf GL}(V)$ since ${\rm Stab}_{{\rm GL}(V)}(0)={\rm Stab}_{{\rm GL}(V)}(V)={\rm GL}(V)$. Actually, it doesn't define a functor at all.
However, to a subspace $W\subset V$ we can also associate a subgroup ${\rm Fix}_{{\rm GL}(V)}(W)\subset{\rm GL}(V)$ consisting of the elements of ${\rm GL}(V)$ that act trivially on $W$. It's easy to see that this rule defines a faithful contravariant functor ${\rm Fix}:{\bf Gr}(V)\rightarrow{\bf GL}(V)$.
3. Evidently, ${\bf Gr}(V)$ is much 'smaller' than ${\bf GL}(V)$. For instance, in the latter there are many finite and discrete subgroups appearing everywhere, and subject to many non-trivial inclusion relations. But still, in the mood of the previous question, is it possible to define some kind of 'adjunction' involving the functor ${\rm Fix}$ ?
Finally, the example I had in mind is the poset ${\bf GL}(V)$. I'm pretty sure it isn't isomorphic to ${\bf GL}(V)^{\bf op}$, but how can one prove that?
I tried to consider, say, its 'almost-initial' objects, i.e., the subgroups $G\subset{\rm GL}(V)$ for which the inclusion $1\rightarrow G$ cannot be non-trivially factorized. These are clearly a lot of discrete subgroups, and possibly some non-discrete ones too, which are loosely related to subspaces of $V$: there's a copy of $S_2=\pm1$ for each subspace, etc. But, how can I pick all of them in such a way that they become comparable to the 'almost-final' objects of ${\bf Gr}(V)$ ? Are there any suggestions of different approaches that might help here?
4. Is the poset ${\bf GL}(V)$ isomorphic to ${\bf GL}(V)^{\bf op}$ ?
Here are answers to your first two questions.
1. If $\phi : V \to V^*$ is any linear map, then for each $u \in V, \phi(u)$ is a linear form on $V$. Therefore $\phi$ is naturally isomorphic to the bilinear form $$\tilde \phi : V \times V \to \Bbb R : (u,v) \mapsto \phi(u)(v)$$ The characteristics of $\tilde \phi$, including signature, are thus applicable to $\phi$.
2. I am assuming that by $m$, you mean the inclusion $m : W \to V : w \mapsto w$. (It is a really bad idea to use non-standard symbolism without introduction.) $\mathbf{Gr}$ is a functor from the category of vector spaces into the category of Posets. If $F \in \operatorname{Hom}(U, V)$, then $$\mathbf{Gr}(F) : \mathbf{Gr}(U) \to \mathbf{Gr}(V) : W \mapsto F(W)$$ is an order-preserving map carrying the subspaces $W$ of $U$ to subspaces of $V$. (Edit: As Qiaochu Yuan points out in the comment, $\mathbf{Gr}(F) : \mathbf{Gr}(V) \to \mathbf{Gr}(U) : W \mapsto F^{-1}(W)$ also defines a contravariant functor. But either way, the op isomorphism will not be natural.)
However, as you defined it, the isomorphism of $\mathbf{Gr}(V)$ with $\mathbf{Gr}(V)^{\bf op}$ is not natural on the category of vector spaces, even if restricted to finite dimensional vector spaces. The reason is that the isomorphism relies on an arbitrarily chosen inner product for its definition. To be natural, a transformation has to rely only on things completely definable from the category properties, not the internal properties of the particular objects under discussion. That inner product is not completely definable simply from the fact that $V$ is a finite-dimensional vector space. We can prove that such inner products exist, but we cannot define a specific one without knowing more about $V$.
That reasoning applies specifically to your definition. It leaves open the possibility that there is an alternate way of defining the same map naturally. But that is not the case. This isomorphism carries $W \mapsto W^\perp$, the space of all vectors perpendicular to $W$ in $V$. Different inner products produce different perpendicular spaces $W^{\perp}$, so the isomorphism between ${\bf Gr}(V)$ and ${\bf Gr}(V)^{\bf op}$ produced does depend on the choice of inner product.
But on the category of inner product spaces, $\bf Gr \to Gr^{op}$ is natural, because this category comes supplied with a particular inner product. This is an example of a general phenomenon: any transformation is "natural" in the right context. You just have to specify enough structure in the categories to allow you to completely define it.