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The cartridge has a 24 ohm impedance. The SUTs have two discrete primary windings which are put in parallel for 50 ohms and in series for 200 ohms. Secondary impedance is 60k ohms. I'm almost 100% sure I'm using the 50 ohms primary.
My understanding is that the cartridge is seeing a 39 ohm load.
OK, the transformer is designed for a 60K load and you are using 47K. That will reduce the Q but only slightly. Assuming (for lack of other evidence) that the design "Q" of the low pass response is 0.707 (a Butterworth filter), your load will slightly reduce the Q, maybe to something closer to a Bessel filter - which should give somewhat better transient response at the price of a rolloff that starts a bit sooner. I doubt you'll hear the difference.
The cartridge sees 50 times (47K/60K) which is 39 ohms. Its output will be reduced by 39/(24+39), a factor of 0.62. The stepup ratio is the square root of (60000/50) or 34.6, so the net stepup is 34.6*0.62 or about 21.5.
If you wanted the cartridge to see a lower impedance, perhaps 24 ohms, you would parallel a 62 ohm resistor across the transformer primary. This would further reduce the output, for a net step up of 17.3.
Does that help?