I am following Lee's Introduction to Smooth Manifolds, but this question is motivated from studying bases of vectors in the context of general relativity. Let $M$ be a smooth $n$-manifold. In the standard construction of a basis of a tangent space, we choose a chart $(U,\varphi)$ whose coordinate domain contains $p\in M$, and the pre-image $(d\varphi_p)^{-1}(\frac{d}{dx^i})\rvert_{\varphi(p)}$ provides a basis on the tangent space $T_pM$ since there is an isomorphism between $T_{\varphi(p)}\mathbb{R}^n$ and $T_pM$.
Since $T_{\varphi(p)}\mathbb{R}^n$ is (isomorphic to) a vector space, I am allowed to choose a different basis for this space. As an explicit example, for $n=3$ in spherical coordinates, the coordinate basis is $\{\partial_r,\partial_\theta,\partial_\phi\}$. But I may wish to redefine my basis to be $\{\partial_r,\frac{1}{r}\partial_\theta,\frac{1}{r\sin(\theta)}\partial_\phi\}$ for every tangent space (Motivation: if I assume the Euclidean metric, this gives me a normalised basis). My question is whether I can find a coordinate map $\psi$ whose coordinate basis is this.
I think my confusion is arising because I haven't actually changed the coordinates used to describe the coordinate domain $U$, but I have changed the basis of the tangent space. The preimage of these new bases are still a basis for $T_pM$ because $d\varphi_p$ is linear. But can my choice of basis at the tangent space induce a new coordinate system?