Why is there no natural isomorphism between $V$ and its dual?

There is an interpretation of natural within category theory that allows us to rigorously state that while there is a natural isomorphism from $V$ to $V''$, there is no natural isomorphism between $V$ and $V'$. This interpretation is explained here and here. The second bit is a little too elaborate for me to wrap my head around, but I'll explain what it is about the $V \to V''$ map which is "natural".

Let $\mathcal C$ denote the category whose objects are finite dimensional vector spaces. The morphisms of this category are the linear maps between vector spaces. We define a functor $F:\mathcal C \to \mathcal C$ by $F(V) = V''$ and $F([V \overset{f}{\to}W]) = [V'' \overset{f''}{\to}W'']$. What makes this a functor is that for any $f:V \to W$ and $g:U \to V$, we have $$ F(f \circ g) = F(f) \circ F(g) $$ We define the much simpler identity functor by $$ \DeclareMathOperator{\id}{id} \id(V) = V; \qquad \id([V \overset{f}{\to}W]) = V \overset{f}{\to}W $$ When we say that $V$ is naturally isomorphic to $V''$, we mean that there is a natural isomorphism between the functors $\id$ and $F$. In this case, what this means is that we can assign an isomorphism (invertible morphism) $\eta_V:\id(V) \to F(V)$ to every vector space $V$ in such a way that:

For every $f:V \to W$, we have $\eta_W \circ \id(f) = F(f) \circ \eta_V$

Or, as we can rephrase it in this context (noting $\id$ is just the identity), we need an $\eta_V:V \to V''$ for every $V$ such that

for any $f:V \to W$, $\eta_W \circ f \circ \eta_V^{-1} = f''$

Now, what is this $\eta_X$? Well, it suffices to take $$ \eta_V:V \to V''\\ [\eta(x)](\alpha) = \alpha(x) $$ You know that this map is an isomorphism from the text. Now, we note that for any $\beta \in V''$, there is an $x_\beta$ for which $\alpha(x_{\beta}) = \beta(\alpha)$ for any $\alpha \in V'$, and we have $\eta^{-1}(\beta) = x_{\beta}$. With that in mind, we can see that for any $f:V \to W$ and for any $\beta \in V''$ and $\alpha \in V'$, we have

$$\begin{align} [[\eta_W\circ f \circ \eta_V^{-1}](\beta)](\alpha) &= [[\eta_W\circ f](x_{\beta})](\alpha) \\ &= [\eta_W(f(x_{\beta}))](\alpha) \\ &= \alpha(f(x_{\beta})) \\ &= [\alpha \circ f](x_{\beta}) \\ &= \beta (\alpha \circ f) \\ &= \beta (f'(\alpha)) \\ &= [\beta \circ f'](\alpha) \\ &= [f'' (\beta)](\alpha) \end{align}$$ as required.