Let $\gamma: V \times V \to K$ be a nondegenerate bilinear form, and let $\overline{\gamma}$ be defined by:
$$\overline{\gamma}: V^* \times V^* \to K, \gamma(x, y) = \overline{\gamma}(\Gamma_\gamma(x), \Gamma_\gamma(y))$$
where $\Gamma_\gamma$ is defined as the linear transformation: $V \to V^*, w \mapsto (v \mapsto \gamma(v, w))$.
Now, if $(V, \gamma$) is an Euclidian Space, I want to show that $(V^*, \overline{\gamma})$ is also an Euclidian Space, and that $\Phi_\gamma: (V, \gamma) \to (V^*, \overline{\gamma})$ is a linear isometry.
I will assume that $V$ has finite dimension. Since $\gamma$ is nondegenerated the map $\Gamma_{\gamma}$ is a isomorphism. Then $\bar{\gamma}$ is, due to your definition, the pull-back of $\gamma$ to $V^*$ through the isomorphism $\Gamma_{\gamma}$ i.e. you defined $\bar{\gamma}$ as $\Gamma_{\gamma}^* \bar{\gamma} = \gamma$. Then $(V^*, \bar{\gamma})$ is an Euclidean space and $\Gamma_{\gamma}$ is (by definition) the isomorphism between $(V,\gamma)$ and $(V^*,\bar{\gamma})$.