 Solution

Yes, the two real smooth vector bundles are isomorphic, although the choice is not canonical. A standard smooth manifold is Hausdorff and second countable, hence admits a smooth locally finite partition of unity. Choose Euclidean Riemannian metrics on coordinate neighborhoods and take their partition-of-unity sum. At each point at least one positive coefficient occurs, so the resulting smooth symmetric tensor $g$ is positive definite.

The resulting musical isomorphism isIt covers the identity of $M$ and is linear on every fibre. In coordinates it is $(x,v)\mapsto (x,G(x)v)$, where $G=(g_{ij})$ is a smooth positive-definite matrix. Its inverse uses the smooth inverse matrix $G^{-1}$, giving the map ${♯}_g$. Thus it is a smooth vector bundle isomorphism.

The Riemannian metric supplies the extra choice: no positive-dimensional real vector space has a linear identification with its dual invariant under all changes of basis. Indeed a scalar change $tI$ acts by $t$ on vectors and $t^{-1}$ on covectors, so an equivariant map $L$ would satisfy $tL(v)=t^{-1}L(v)$ for every $t>0$, forcing $L=0$. An isomorphism therefore exists without being a metric-free canonical one.