Here is an Example in Euclidean 3-space: When using spherical coordinates $(r, \theta, \phi)$ with $\theta$ and $\phi$ the polar and azimuthal angles, respectively: a natural basis for these coordinates given the metric 
$$(g\_{ij}) = \begin{pmatrix}
1 & 0 & 0 \newline
0 & r^2 & 0 \newline
0 & 0 & r^2 \sin^2{\theta}
\end{pmatrix}$$
is $$\vec{e\_{r}} = \begin{pmatrix}
1 \\ 0 \\ 0
\end{pmatrix}, \\
\vec{e\_\theta} = \begin{pmatrix}
0 \\ 1 \\ 0
\end{pmatrix}, \\ 
\vec{e\_\phi} = \begin{pmatrix}
0 \\ 0 \\ 1 
\end{pmatrix}$$
which have the following lengths:
$$||\vec{e\_{r}}|| = 1, ||\vec{e\_{\theta}}|| =r, ||\vec{e\_{\phi}}|| = r \sin{\theta}.$$


Similarly for the analogously defined dual basis:
$$||\vec{e^{r}}|| = 1, ||\vec{e^{\theta}}|| =\frac{1}{r}, ||\vec{e^{\phi}}|| = \frac{1}{r \sin{\theta}}.$$


In physics it is common to define a new, unit basis (hence-force the hatted basis): 
$$\vec{e\_{r}} = \hat{e\_r}, \\ \\ \\ \\ \frac{\vec{e\_{\theta}}}{r} =\hat{e\_\theta}, \\ \\ \\ \\ \frac{\vec{e\_{\phi}}}{r \sin{\theta}} = \hat{e\_\phi}$$


The question then arises as to what the dual basis is for this normalized coordinate system, labelled $\hat{e^r}, \hat{e^\theta}, \hat{e^\phi}$ (hats for consistency, but not necessarily norm of $1$): $$\hat{e^\theta} = g^{\theta \theta}\hat{e\_\theta} = g^{\theta \theta}\frac{\vec{e\_\theta}}{r} = \frac{\vec{e^\theta}}{r}, || \hat{e^\theta}|| = \frac{1}{r^2}$$
$$\hat{e^\phi} = g^{\phi \phi}\hat{e\_\phi} = g^{\phi \phi}\frac{\vec{e\_\phi}}{r \sin{\theta}} = \frac{\vec{e^\phi}}{r \sin{\theta}}, || \hat{e^\phi}|| = \frac{1}{r^2 \sin^2{\theta}}$$


Consider the vector in the natural basis $$\vec{V} = \begin{pmatrix}
a \\ \\ b \\ \\ c
\end{pmatrix} .$$ This same vector in the normalized basis is:
$$\hat{V} = \begin{pmatrix}
a \\ \\ br \\ \\ cr \sin(\theta) 
\end{pmatrix} .$$


Given that a metric is defined by $g\_{ij} = \langle e\_i,e\_j \rangle $ we can see our change of basis has preserved the inner product, i.e. $\vec{V}^2 = \hat{V}^2$ with our new metric, $$(\hat{g\_{ij}}) = (\delta\_{ij}) = \begin{pmatrix}
1 & 0 & 0 \newline
0 & 1 & 0 \newline
0 & 0 & 1 
\end{pmatrix}.$$ However if we do this analysis with the dual basis we do not find the inverse metric, but another metric: $$(\hat{g^{ij}}) = \begin{pmatrix}
1 & 0 & 0 \newline
0 & \frac{1}{r^4} & 0 \newline
0 & 0 & \frac{1}{r^4 \sin^4{\theta}}
\end{pmatrix} \neq (\delta^{ij})$$


This set of $2$ not inverse metrics for vectors and co-vectors respectively, is the crux of my confusion. Is this somehow okay? If not, is there something I have done incorrectly here, computationally or definitionally? My question more broadly is: how does going from a natural basis which is implied from the metric, to a normalized vector basis, in general affect the induced dual space and metric? 



What happens to the metric when we normalize the basis?
The metric tensor is defined from the basis, $g\_{ij} = \vec{e}\_{i} \cdot \vec{e}\_{j} $. If you change the basis from the $\vec{e}\_{i} $ to the $\hat{e}\_{i} $, then you also change the metric tensor from $g\_{ij} $ to the corresponding $\hat{g}\_{ij} $. You can't calculate the new $\hat{e}^{i} $ using the old metric $g^{ij} $, as you do above; instead, it has to be $\hat{e}^{i} = \hat{g}^{ij} \hat{e}\_{j} $. The dual basis should satisfy $\hat{e}^{i} \cdot \hat{e}\_{j} = \delta^{i}\_{j} $.


