= Solution
\b[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> is
$$
\boxed{\flat_g:TM\longrightarrow T^*M,\qquad v\longmapsto g_p(v,\cdot).}
$$
It 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 $\sharp_g$. 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.
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