**Edit:** According to interesting comment of Thomas Rot to the previous version of the question, we revise the question as follows:

First note that if a manifold $M$ is a parallelizable manifold , then it gets a natural Riemannian metric which is independent of the base point $x\in M$.In fact $TM \simeq M \times \mathbb{R}^n$ enable us to carry the standard Euclidean inner product to each fiber of $TM$. In the following question we apply this obvious fact to $M=TS^n$ as follows:

It is well known that the tangent bundle $TS^n$ of $S^n$ is a parallelizable manifold, then once we fix a trivialization for its tangent bundle(The tangent bundle of $TS^n$), $TS^n$ gets s a natural Riemannian metric, the Euclidean one along each fiber of its tangent bundle. Now we can restrict this metric on $T S^n$ to the zero section $S^n$.

Is there a trivialization of $TS^n$ for which the corresponding restricted metric coincide to the standard metric of $S^n$?