PretentiousFood:
I'm not quite sure how you have hooked up the FET to the tube tracer. If by degeneration resistor you mean a resistor going from source to ground, but you connected the gate stopper resistor to grid voltage rather than the bottom end of the source resistor, then this is not going to give accurate results for the impedance of a current source. The reason is that in a FET current source, the gate has to be connected to the bottom end of the source resistor. - see the current sources in the SRX schematic. This results in a negative feedback loop that increases the impedance, and also sets the static current.
Here is how it works: suppose we try to increase the current running through the FET. The increase in the current increases the voltage across the source resistor, which in turn means that the gate voltage becomes relatively more negative which tends to turn off the FET, making the current decrease.
The correct way to see its behavior as a current source is to treat it as a two-terminal device rather than a three-terminal device, which is what a tube tracer is usually doing. Instead of plugging in the FET like a tube, you have to use both gate and source resistors, and connect the gate resistor to the bottom end of the source resistor, then run the I vs V curves - you should get a dV/dI in the hundreds of kilohms, which is what both Pimm and Jung measured. The gate should NOT be connected to a constant grid voltage, which appears to be how you did the measurement. Jung specifically stated his measurements of the DN2540 were made at around 30 mA static current, where he measured about 300 kilohms, which is about 10x what you measured. If what I suggest is in fact the way you did the measurement then I apologize for the misunderstanding, however your results are so far off from his and Pimm's I suspect that that is not how you did it.
To go one step further, the reason a cascode current source works so well is that a large variation in the drain-to-source voltage of the upper device causes relatively little change in the gate-to-source voltage of that device. Almost all of the variation occurs in the drain-to-gate voltage. The drain and source of the lower device in the cascode is connected between the gate and source of the upper device, therefore its drain-to-source voltage is nearly constant even while there is a wide variation in the voltage across the two device combination. Since the lower device controls the current, and since it is running at close to a constant voltage regardless of voltage variations across the cascode, and the lower device is also getting the negative feedback from the source resistor as described above, it should output a constant current - or very very close to it. And that is the definition of a current source. With the cascode, you should get a dV/dI on the order of a hundred megohms or more.