= Solution
For a <bilinear form> $\phi$, define $F_\phi:V\to V^*$ by $F_\phi(v)(u)=\phi(u,v)$. Bilinearity makes $F_\phi(v)$ a linear functional and makes $F_\phi$ linear in $v$. Conversely, every linear $F:V\to V^*$ defines a bilinear form by $\phi_F(u,v)=F(v)(u)$. These constructions undo each other, giving the desired bijective correspondence.
For a <nondegenerate bilinear form>, $F_\phi$ has zero kernel and is an isomorphism because $V$ and its <dual space> have the same finite dimension. For the two forms, put
$$
\boxed{\alpha=F_{\phi_1}^{-1}F_{\phi_2}.}
$$
It is an isomorphism and satisfies $\phi_1(u,\alpha v)=F_{\phi_1}(\alpha v)(u)=F_{\phi_2}(v)(u)=\phi_2(u,v)$ for all $u,v$. This is the <representation of a bilinear form relative to a nondegenerate bilinear form>.
If both forms are symmetric, then
$$
\phi_1(u,\alpha v)=\phi_2(u,v)=\phi_2(v,u)=\phi_1(v,\alpha u)=\phi_1(\alpha u,v).
$$
Nondegeneracy makes the adjoint with respect to $\phi_1$ unique; this equality proves \b[$\boxed{\alpha^\dagger=\alpha}$] in that bilinear-form sense. No positivity or Hermitian conjugation is assumed over the arbitrary field.
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