Let $M$ be a smooth manifold. Pick some Riemannian metric $g_0$ on $M$, and let $G := \mathrm{Isom}(M,g_0) \subset \mathrm{Diff}(M)$.
Now, by classifying space theory, for a manifold $X$ we have an equivalence of groupoids $$ (\text{principal $G$-bundles on $X$}) \simeq [X,BG] \,, $$ where in the RHS this is the groupoid with objects maps $X\rightarrow BG$ and morphisms the homotopies between these.
However, do we not also have an equivalence of groupoids $$ (\text{principal $G$-bundles on $X$}) \simeq (\text{$M$-fiber bundles over $X$ equipped with fiberwise Riemannian metric}) $$ given in the forward direction by taking the associated bundle w.r.t. the $G$ action on $M$? (Analogously with how principal $O(n,\mathbb{R})$-bundles are equivalent to rank-$n$ real vector bundles with bundle metric.)
Thus we should have an equivalence between $M$-fiber bundles on $X$ equipped with fiberwise Riemannian metric, and maps $X \rightarrow BG$, up to "coherent homotopy".
In the case of $X = \mathrm{pt}$, an $M$-fiber bundle over a point with a fiberwise Riemannian metric, is just a choice of Riemannian metric on $M$.
Thus the groupoid of Riemannian metrics on $M$ is equivalent to the groupoid $[\mathrm{pt},BG]$; more simply, the space of Riemannian metrics on $M$ is $BG$.
My first question is, do the steps above make sense? Especially the equivalence $$ (\text{principal $G$-bundles on $X$}) \simeq (\text{$M$-fiber bundles over $X$ equipped with fiberwise Riemannian metric}) $$ in the 2nd bullet point?
Next, couldn't we also treat the space of Riemannian structures on $M$ as the space of lifts $\tilde{f} : M\rightarrow BO_m$ of $M$'s tangent bundle classifying map $M \rightarrow BGL_m$?
Is there some direct way to see how these two models of the "moduli space of Riemannian metrics" coincide?