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?
