Why is there no natural isomorphism between $V$ and its dual? Question
While looking over the exercise $3.F-34$ in Linear Algebra Done Right, I encountered the following paragraph

Suppose $V$ is finite dimensional. Then $V$ and $V'$ are isomorphic, but finding an isomorphism from $V$ onto $V'$, generally, requires choosing a basis of $V$. In contrast, the isomorphism from $V$ to $V''$ does not require a choice of basis and thus is considered more natural.

and these questions showed up in my mind:
$1$. Does the word natural just means that we don't need to choose a basis? I have seen the word canonical is used in the similar manner too. Is there a more precise definition for natural or canonical?
$2$. Assuming the answer to question $1$ is Yes, then why there is no natural isomorphism from $V$ onto $V'$?
$3$. I think that there is a relation between the answer to question $2$ and the proof of Riesz representation theorem. So, if we cannot find a natural isomorphism between $V$ and $V'$ then it means that we cannot prove Riesz representation theorem without choosing a basis of $V$. Is this true?

Complementary Informations
The Isomorphism from $V$ onto $V^{''}$.
Suppose $V$ is a finite dimensional vector space. Consider the following map
$$
\Lambda(v)(\phi)=\phi(v),  \qquad \forall v \in V, \,\, \forall \phi \in V^{'}
$$
then $\Lambda$ is an isomorphism from $V$ onto $V''$.
Riesz Representation Theorem.
Suppose $V$ is a finite dimensional linear space equipped with an inner product and $\phi$ is a linear functional on $V$. Then there is a unique vector $v_0 \in V$ such that
$$\phi(v) = {\langle v,v_0 \rangle}_{V}, \qquad \forall v \in V$$

Other Related Posts
I found the following posts related to this question on MSE and MO.
Post $1$, Post $2$, Post $3$, Post $4$.
 A: 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.
