Let $(M,g)$ be a $(n+1)$-dimensional Riemannian manifold. Let $\{\alpha^0,\ldots,\alpha^n\}$ be a local orthonormal coframe, so that the metric $g$ locally reads $g = \sum_{j=0}^n \alpha^j\otimes \alpha^j$. The structure equations in this coframe are \begin{align} d\alpha^j &= - \omega^j_k\wedge \alpha^k, & \forall j &\in \{0,\ldots,n\} \\ 0 &= \omega^i_j + \omega^j_i, & \forall i,j&\in \{0,\ldots,n\}. \end{align} I am trying to derive the structure coefficients of an inhomogeneous scaling of the metric. More precisely, for $A$ and $B$ two positive constants, consider $\bar{\alpha}^0 = A \alpha^0$ and $\bar{\alpha}^j = B \alpha^j$ for $j\in \{1,\ldots,n\}$. Let $\bar{g} = \sum_{j=0}^n \bar{\alpha}^j\otimes\bar{\alpha}^j$. What are the structure coefficients $\bar{\omega}^i_j$ of $\bar{g}$ in the orthonormal coframe $\{\bar{\alpha}^0,\ldots,\bar{\alpha}^n\}$?
A quick inspection shows that \begin{align} d\bar{\alpha}^0 &= -\omega^0_0\wedge \bar{\alpha}^0 - \sum_{k=1}^n \frac{A}{B} \omega^0_k\wedge\bar{\alpha}^k, \\ d\bar{\alpha}^j &= -\frac{B}{A}\omega^j_0\wedge \bar{\alpha}^0 - \sum_{k=1}^n \omega^j_k\wedge \bar{\alpha}^k, & \forall j \in \{1,\ldots,n\}. \end{align} It is tempting to conclude that $\bar{\omega}^0_j = \frac{A}{B} \omega^0_j$ and $\bar{\omega}^j_0 = \frac{B}{A}\omega^j_0$ for $j\in \{1,\ldots,n\}$. But this cannot be true: this won't make the coefficients $\bar{\omega}^i_j$ skew symmetric as soon as $A \neq B$.
My guess is that $\bar{\omega}^0_j = \frac{A}{B}\omega^0_j + f^0_j \bar{\alpha}^0$ and $\bar{\omega}^j_0 = \frac{B}{A}\omega^j_0 + f^j_0\bar{\alpha}^j$, with $\{f^i_j\}$ some functions (presumably constant functions). I couldn't manage to go further than that yet.
Of course, my goal is to derive the curvature of $\bar{g}$ in terms of the curvature of $g$.
Let $g$ be your original metric, and $\{e_i\}$ an orthonormal frame. Then define the coefficients by: \begin{align*} [e_a,e_b]=\Omega_{ab}^ce_c \end{align*} Furthermore, in this local frame write the Levi-Civita connection as: \begin{align*} \nabla e_a=\omega_a^b\otimes e_b \end{align*} for one forms $\omega_a^b$. The metric compatibility of $\nabla$ forces these one forms to be skew symmetric in the indices $a$ and $b$. We note that: \begin{align*} g(e_a,e_b)=\delta_{ab} \end{align*} implying that the first three terms in the Koszul equation vanish. Furthermore, we have that: \begin{align*} g([e_c,e_a],e_b)=g(\Omega_{ca}^de_d,e_b)=\Omega_{ca}^d\delta_{db}=\Omega^b_{ca} \end{align*} Furthermore: \begin{align*} g(\nabla_{e_c}e_a,e_b)=g(\omega_a^d(e_c)e_d,e_b)=\omega_a^d(e_c)\delta_{db}=\omega_a^b(e_c) \end{align*} The Koszul equations then implies that: \begin{align*} \omega_a^b(e_c)=\frac{1}{2}\left(\Omega_{ca}^b-\Omega_{ab}^c+\Omega_{bc}^a\right) \end{align*} It follows that if $\{\bar{e}_0,\dots ,\bar{e}_{n+1}\}=\{A_0^{-1}e_0,\dots, A_{n+1}^{-1}e_n\}$, then the new connection one form is defined by: \begin{align*} \bar{\omega}^b_a(\bar{e}_c)=\frac{1}{2}\left( \bar{\Omega}_{ca}^b-\bar{\Omega}_{ab}^c+\bar{\Omega}_{bc}^a\right) \end{align*} where: \begin{align*} [\bar{e}_a,\bar{e}_b]=\bar{\Omega}^c_{ab}\bar{e}_c \end{align*} Note that: \begin{align*} [\bar{e}_a,\bar{e}_b]=A_a^{-1}A_b^{-1}[e_a,e_b]=A_a^{-1}A_b^{-1}\Omega_{ab}^ce_c \end{align*} so: \begin{align*} A_a^{-1}A_b^{-1}\Omega_{ab}^ce_c=A_c^{-1}\bar{\Omega}^c_{ab}e_c \end{align*} implying that: \begin{align*} \bar{\Omega}^c_{ab}=A_cA_a^{-1}A_b^{-1}\Omega_{ab}^c \end{align*} hence: \begin{align*} \bar{\omega}_a^b(\bar{e}_c)=\frac{1}{2}\left(A_c^{-1}A_a^{-1}A_b\Omega_{ca}^b -A_a^{-1}A_b^{-1}A_c\Omega_{ab}^c +A_b^{-1}A_c^{-1}A_a\Omega_{bc}^a\right) \end{align*} This is about the best I can do, but maybe some ugly algebra can help you rewrite this in terms of $\omega_a^b$.