Let $(M,g)$ be a Riemannian manifold. Then there is the well-known
Sasaki metric that makes $(TM,\hat{g})$ a Riemannian manifold. In a
similar way, one can construct a Sasaki metric $\bar{g}$ on the
cotangent bundle. My question is, is the map
$g\colon (TM,\hat{g}) \to (T^\*M,\bar{g})$ an isometry?


For completeness, the Sasaki metrics are given as follows (not 100%
sure about the cotangent one). Let $X,Y$ be vector fields on $M$ and
$\alpha,\beta$ be one-forms on $M$, let $\pi$ be the projection from
both bundles onto $M$ and let superscripts $h$ and $v$ denote
horizontal and vertical lifts. Then these Sasaki metrics are uniquely
defined by the following conditions:
$$
\hat{g}(X^h,Y^h) = \hat{g}(X^v,Y^v) = g(X,Y) \circ \pi, \quad \hat{g}(X^h,Y^v) = 0
$$
and
$$
\bar{g}(X^h,Y^h) = g(X,Y) \circ \pi, \quad
\bar{g}(\alpha^v,\beta^v) = g^{-1}(\alpha,\beta) \circ \pi, \quad
\bar{g}(X^h,\alpha^v) = 0.
$$



Are the Sasaki metrics on tangent and cotangent bundle isomorphic?
The answer is yes, and you can see this by realizing that the vertical (resp. the horizontal) subbundle of $TTM$ is mapped to the vertical (resp. the horizontal) subbundle of $TT^\*M.$ For the vertical part, this follows from the fact that $g\colon TM\to T^\*M$ is a bundle homomorphism, and for the horizontal bundle this follows from the fact that $g\colon TM\to T^\*M$ is parallel (with respect to the natural connections defined by $g$ which are used to define the horizontal bundles).


