The vielbein postulate is given by
$$\partial_\mu e_\nu{}^a + \omega_\mu{}^a{}_b e_\nu{}^b - \Gamma_{\mu\nu}^\rho e_\rho^a = 0 $$
whose anti-symmetric part, assuming the connection is a Levi-Civita connection, reads
$$2\partial_{[\mu} e_{\nu]} - 2\omega_{[\mu}{}^{ab}{} e_{\nu ]b} = 0 $$
The solution of this equation gives rise to an equation for the spin-connection $\omega_\mu{}^{ab}$ in terms of $e_\mu{}^{a}$.
As we use the "vielbein postulate", or the antisymmetric part of the vielbein postulate, we must be careful with the variation of this postulate. The variation of the postulate must also vanish otherwise it would yield a constraint. The transformation rules for the vielbein and the spin connection are
$$ \begin{align} \delta e_\mu{}^a &= \partial_\mu \xi^a + \omega_\mu{}^{ab} \xi_b - \lambda^{ab} e_{\mu b}\\ \delta \omega_\mu{}^{ab} &= \partial_\mu \lambda^{ab} + \omega_{\mu c}{}^{[a} \lambda^{b]c} \,. \end{align} $$
where $\xi_a$ and $\lambda_{ab}$ are the transformation parameters for boosts and Lorentz transformations. The transformation of the antisymmetrized vielbein postulate then reads
$$\xi_\nu R_{\rho\sigma}{}^{\mu\nu} = 0$$
provided that the vielbein postulate is satisfied. Here $R_{\mu\nu\rho\sigma}$ is the Riemann tensor. I have never seen such an identity. My question is: Has anyone ever seen something like this? Or am I missing something?