This is not an answer, but rather a long comment. Software like Mathematica can plot trajectories of this dynamical system, by (numerically) solving the system of ODEs. Here are trajectories that start on a circle of radius $1/10$ centered around the lowest nontrivial zero of $\zeta(s)$ (sorry, I can't make myself call the variable $Z$.) [![enter image description here][1]][1]
The picture near the zero is exactly what you expect when you think of approximating $\zeta(s)$ by the linear approximation: the ODE in $s=\sigma+it$ is very nearly $\sigma^\prime=\sigma$ and $t^\prime=t$, at least, up to the scaling and rotation of the picture by $\zeta^\prime$ evaluated at the zero. Further from the zero, higher order terms come in to play.
Here are trajectories that start on a (larger) circle of radius $1/2$ centered on the pole at $s=1$. This picture is also what you should expects when you think about the Laurent expansion of $\zeta(s)$ at $s=1$
[![enter image description here][2]][2]
Update (Things I should have remembered from complex analysis in my original posting): The function $f(s)=(s-1)\zeta(s)$ is holomorphic in $\mathbb C\backslash \{1\}$. Let $u(\sigma,t)$ and $v(\sigma,t)$ be the real and imaginary parts, as functions of $s=\sigma+i t$ (NB: $t$ is not time in this notation.) From the point of view of complex analysis, it makes more sense to look instead at the Polya vector field $(u,-v)$ , which is conservative. Indeed, with $$ F(s)=\int_{s_0}^s f(w)\, dw=U(\sigma,t)+iV(\sigma,t) $$ we have $\triangledown U=(U_x,U_y)=(U_x, -V_x)=(u,-v)$ The trajectories for this vector field are the level curves of $V$.
I don't believe there are any closed orbits. [1]: https://i.sstatic.net/xWsxN.jpg [2]: https://i.sstatic.net/6LQMI.jpg
