Yes, this is possible, but the word "adiabatic" is not usually used to describe this. I will call a process that you describe a no-external-heat-flow process, to distinguish from an adiabatic, or constant entropy, process.
The only restriction is that at at any two successive stopping points, the entropy must go up. To realize this, first do a true adiabatic transformation to get you to the volume you want, then quickly jiggle the piston, making soundwaves that dissipate heat into the gas until you get it to the temperature you want. The combinations of adiabatic lines and heat-generating volume-preserving lines let you approximate any curve that links points such that successive points are always higher entropy.
The reason why you arrive at different solutions is that the assumptions in the assignment are inconsistent. One way to show this is to show in isothermal flow if ideal gas in straight duct of constant cross-section with no friction the gas has to have the same pressure everywhere; different pressures at the entrance and the exit $P_1, P_2$ are not possible.
To show this, let us take the Euler equation
$$\
\rho v \frac{dv}{dx} + \frac{dp}{dx} = 0,
$$
and transform it into equation for $P(x)$ only. We express $v$ as $G/\rho$ and obtain
$$
\frac{d}{dx}\left(\frac{G^2}{\rho}\right) + \frac{dP}{dx} = 0.
$$
From the state equation of ideal gas, we can express the density as a function of pressure and temperature:
$$
\rho = \frac{MP}{RT},
$$
where $M$ is molar mass of the gas and $R$ is the universal gas constant. Replacing $\rho$ in the last equation, we obtain
$$
\frac{d}{dx}\left(\frac{G^2 RT}{M} \frac{1}{P(x)} + P(x)\right) = 0.
$$
Since the temperature $T$ is assumed constant, this equation implies that the pressure $P$ is constant as well, which contradicts the assumption about different pressures $P_1, P_2$.
In other words, if there is no friction, the gas will move inertially with the same pressure and velocity everywhere. The velocity can be any number so it cannot be deduced from the data in the assignment.
The assignment may have reasonable solution if we introduce friction into the model. The mathematically simplest way seems to be to add a negative constant force to the right-hand side of the Euler equation:
$$
\frac{d}{dx}\left(\frac{G^2 RT}{M} \frac{1}{P(x)} + P(x)\right) = -f.
$$
This equation has sensible solution $P(x)$ for given pressures $P_1, P_2$ and will allow you to find $G$ and $v(x)$. However, physically constant friction is rather unrealistic model, especially if the friction is low, as then the gas velocity and its transversal gradient will increase rapidly along the duct. Since the Newton friction forces are proportional to transversal gradient, they should get stronger along the duct and thus damp the increase in velocity (but not completely, in reality the velocity should increase along the duct). More realistic approach would be to solve the Navier-Stokes equations for stationary isothermal flow, similarly to what one does in the derivation of the Poiseuille law for incompressible liquid flow in a pipe.
Best Answer
Yes, after a fashion, increase of entropy is always coupled to some kind of "friction". Though isentropic could be argued to be stronger than frictionless, as it does not only refer to mechanical friction. For example a Rayleigh flow with heat transfer will not be isentropic (credits to Azad for pointing out the example).
Nearly, it is $v_2 = \sqrt{v_1^2 - 2g \Delta z}$. But note, that constant temperature does not necessarily imply isentropicity.