Uniqueness of the evaluation isomorphism as a natural isomorphism It is well known that the evaluation map between $V$, a topological vector space, and $V^{\star\star}$, the double dual, given by $\epsilon_{v} = f\mapsto f(v)$ is a natural isomorphism between the identity functor and the functor $(-)^{\star\star}$ in the category of real finite dimensional vector spaces or in the category of reflexive Banach spaces.
Someone once, perhaps mistakenly, told me that a topological vector space is reflexive it is naturally isomorphic to its double dual. The definition that I recall is that such a space is reflexive if the evaluation map is an isomorphism. I want to know if the previous definition is definitely wrong in a certain sense: Is it possible that for some sub-category of topological vector spaces, there is a natural isomorphism between the identity functor and the double dual functor that does not correspond to the evaluation map?
 A: I initially posted the following, but deleted it because I don't know enough about topological vector spaces. Since nobody posted an answer, I'll undelete it. Even though the comment about evaluation at $-v$ answers the question as it is asked, it doesn't invalidate the imprecise definition given to the OP. This definition was probably a mistake but some interpretation of it is still valid.

First, there are Banach spaces isomorphic to their double dual without being reflexive. It is the subject of James' 1951 article A non-reflexive Banach space isometric with its second conjugate space.
But the erroneous definition says a space $V$ is reflexive if it is naturally isomorphic to its double dual. We will take this as meaning that there is an isomorphism $V→V^{**}$ natural with respect to all the endomorphisms of $V$. And we can prove that such a natural morphism is necessarily a multiple of the evaluation map, so if it's an isomorphism, the evaluation map too. Here is a proof. It uses the Hahn-Banach theorem, so I have no idea how badly things can go with general topological vector spaces.
Let $V$ be a Banach space and let $f : V→V^{**}\newcommand{\ev}{\operatorname{ev}}$ be a map natural in the endomorphisms of $V$. Choose $x ∈ V$ different from $0$ and choose some projector $p : V → ℝx$ which is the identity on $x$. We have $p(y) = w(y)x$ for some $w ∈ V^*$, so
$$p^*(u) = u∘p = u(x) w = \ev_x(u) w\text{.}$$
We deduce that $p^*$ is a projector on $ℝw$ and reapplying the same reasoning, $p^{**}$ is a projector on $\ev_x$ (explicitly, $p^{**}(h) = w^*(h) \ev_x$).
Since $p^{**}$ is a projector on $ℝ \ev_x$, we must have $f(x) = α \ev_x$ for some $α∈ℝ$, because $p^{**}(f(x)) = f(p(x)) = f(x)$. The goal is to prove that this $α$ is the same for all $x ∈ V$. In order to do that, let $y ∈ V$ and take $r : V→V$ defined by $r(z) = w(z) y$. Then $f(y) = f(r(x)) = r^{**}(f(x)) = r^{**}(α \ev_x) = α \ev_y$.
But I don't have very clear ideas about these duals of duals. It should be possible to explain that better...
