By differentiating $\xi$ and solving for $\zeta(1-x)$:

$$ \zeta(1-x) = \frac{2(\zeta'(x)\Gamma(x/2)+\Gamma((1-x)/2) \zeta'(1-x)\pi^{x-1/2}) )}{\Gamma((1-x)/2) \pi^{-1/2+x}(2\log\pi -\psi((1-x)/2)-\psi(x/2))} \qquad (1)$$

and unless the denominator vanishes (like near $\frac12 \pm 6 i$) the non-trivial zeros of zeta are the zeros of

$$\zeta'(x)\Gamma(x/2)+\Gamma((1-x)/2) \zeta'(1-x)\pi^{x-1/2} = 0$$

Is something similar possible for Riemann $\xi$ or symmetrized zeta $\zeta^*$?