Inner products on vector spaces and identification of a space with its dual We know that given an inner product on a finite dimensional vector space $V$, we can find an identification (isomorphism) of $V$ with $V^*$ (dual space) as follows : 
 Take any orthonormal basis w.r.t the inner product , and map this basis to the dual basis (respectively for each basis element). This isomorphism does not depend on which orthonormal basis was chosen. 
My question is : 

Is every isomorphism $\alpha : V \to V^* $ obtained by some choice of inner product on $V$ ?
  If not, given an isomorphism $\alpha$, how do we test whether it is obtainable in this manner from some choice of metric ? Is there some neat characterization of such isomorphisms ?

My guess is that the asnwer to the first question is in the negative. Intuition : There are about $GL(n;\mathbb{R})$ many isomorphisms with $n^2$ independent entries, but less than $n(n+1)/2$ different choices of metrics. But I am looking for a more precise answer.
 A: Let $(V, \left< \cdot, \cdot \right>)$ be an $n$-dimensional real inner product space and write $g(v,w) = \left<v, w \right>$. Choose an orthonormal basis $(v_1, \dots, v_n)$ and denote by $(v^1, \dots, v^n)$ the corresponding dual basis of $V^{*}$. Let $\alpha_g \colon V \rightarrow V^{*}$ be the isomorphism uniquely defined by $\alpha_g(v_i) = v^i$. Note that given $v,w \in V$ we have
$$ \alpha_g(v)(w) = \alpha_g \left( \sum_{i=1}^n \left<v, v_i \right> v_i \right)(w) = \sum_{i=1}^n \left<v, v_i \right> \alpha_g(v_i)(w) = \sum_{i=1}^n \left< v, v_i \right> v^i(w) = \sum_{i=1}^n \left< v, v_i \right> \left< w, v_i \right> = \left< \sum_{i=1}^n \left< v, v_i \right> v_i, \sum_{j=1}^n \left< w, v_j \right> v_j \right> = \left< v, w \right>. $$
This implies that you can reconstruct the inner product $g$ from the isomorphism $\alpha_g$ and places a restriction on the possible $\alpha$'s that can be obtained by the construction above.
Going the other direction, given an isomorphism $\alpha \colon V \rightarrow V^{*}$, you can define a bilinear form on $V$ by
$$ (v,w)_{\alpha} := \alpha(v)(w). $$
What we shown above is that a necessary condition for $\alpha$ to come from some metric $g$ is that  $(\cdot, \cdot)_{\alpha}$ must be an inner product (symmetric and positive-definite). You can check that this condition is also sufficient by defining $g(v,w) = \left(v, w \right)_{\alpha}$ and then checking that $\alpha_g = \alpha$.
A: The answer is, like you guessed, no. An example is given by taking an arbitrary finite dimensional vector space $V$ over some field $K$ with no inner product, with basis $\{e_1, ..., e_n\}$. Then let $\alpha : V \rightarrow V^*$ be given by $\alpha(e_i) = \varphi_{e_i}$ and expanding linearly, where $\varphi_{e_i}$ is given by $\varphi_{e_i}(e_j) = \delta_{ij}$ and expanding linearly. So the concept of dual basis makes sense without an inner product structure. However, it is true that one can define an inner product for any isomorphism $ \alpha : V \rightarrow V^*$ of the above form such that the isomorphism induced by that inner product is $\alpha$. If $\{e_1, ... , e_n \}$ is the chosen basis, just define: 
$$<\sum_i \lambda_i e_i, \sum_i \mu_i e_i> = \sum_i \lambda_i \mu_i.$$
