I'm going to assume that "pushed towards" means that the magnet is given some initial velocity towards the tube (starting at some large distance away) and then no further forces act on it. I'll also assume that everything is nice & non-relativistic, and that any energy or momentum lost to radiation is negligible.
The force between the magnet and the tube will be proportional to their relative velocity $v_\text{rel}$. To see this, note that in the instantaneous reference frame of the tube, we have $\vec{\nabla} \times \vec{E} = - \partial \vec{B}/\partial t$, and $\partial B/\partial t$ is proportional to $v_\text{rel}$. Thus, the electric field due to the motion of the magnet is also proportional to $v_\text{rel}$, the current density in the tube is proportional to $v_\text{rel}$, and so the force on the current due to the magnet is also proportional to $v_\text{rel}$. The exact proportionality factor is tricky to calculate, but we can definitely say that $F = - \alpha v_\text{rel}$, where $\alpha$ depends on the physical properties of the system such as the strength of the magnet, the thickness & material of the pipe, etc.
This means that the equation of motion for the magnet will be
$$
m_m \ddot{x}_m = - \alpha (\dot{x}_m - \dot{x}_p)
$$
and that for the pipe will be
$$
m_p \ddot{x}_p = -\alpha (\dot{x}_p - \dot{x}_m)
$$
respectively. (The subscript $m$ denotes the magnet and the subscript $p$ denotes the pipe.) Note that the two forces are equal & opposite, as they must be if we want to have conservation of momentum in this problem.
Defining $x_{cm}= (m_m x_m + m_p x_p)/(m_m + m_p)$ to be the position of the center of mass and $\delta = x_m - x_p$ to be the relative displacement in position, it can be shown that the two equations above are
equivalent to
$$
\ddot{x}_{cm} = 0
$$
and
$$
\mu \ddot{\delta} = -\alpha \dot{\delta},
$$
where $\mu$ is the reduced mass defined by $1/\mu = 1/m_m + 1/m_p$. The first equation says that the center of mass of the system moves at constant velocity; if we assume that the initial velocity of the pipe is zero and denote the initial velocity of the magnet as $v_{m0}$, then we have
$$
\dot{x}_{cm} = \frac{m_m}{m_m + m_p} v_{m0}.
$$
The second equation says that the relative velocity decays exponentially with time:
$$
\dot{\delta} = v_{m0} e^{-\alpha t/\mu},
$$
This allows us to rewrite the velocities of the magnet and pipe as functions of $t$:
$$
v_{m} = \dot{x}_{cm} + \frac{m_p}{m_m + m_p} \delta = v_{m0} \frac{m_m + m_p e^{-\alpha t/\mu}}{m_m + m_p}
$$
and
$$
v_p = \dot{x}_{cm} - \frac{m_m}{m_m + m_p} \delta = v_{m0} \frac{m_m(1 - e^{-\alpha t/\mu})}{m_m + m_p}
$$
Physically, what happens is that the eddy current forces would transfer some momentum from the magnet into the pipe. Precisely how quickly this momentum transfer takes place depends on all the physical properties of the system encoded in the constant $\alpha$, but for late times $t \gg \mu/\alpha$ (assuming the magnet has not exited the pipe by this time), the relative velocity will become negligibly small, and the pipe and the magnet will be moving in tandem with speed $v \approx m_m v_{m0}/(m_m + m_p)$. In other words, if the pipe is long enough, the whole setup acts like an inelastic collision between the pipe and the magnet.
Finally, note that the above is done for the general case (Case #1 in the OP.) Case #2 can be obtained as a special case by assuming that $m_p \gg m_m$, in which case we find that $\dot{x}_m \approx \dot{\delta} \approx v_{m0} e^{-\alpha t/m_m} $ ($\mu \approx m_m$ in this limit.) In other words, the magnet's speed decays to zero exponentially, and the pipe (and space station) barely move.
