What has always boggled my mind is that SUT manufacturers so rarely publish the turns ratio(s) of their products. Which would make it very simple indeed to at least begin to match a SUT to a particular cartridge. Instead, and especially in the golden era of vinyl, most SUTs were given a value in ohms (see Elliot’s compendium above). Typically, the value in ohms is meant to indicate the recommended internal impedance of the cartridge to which you want to match the SUT. For example a "3 ohm" SUT is meant to be used with cartridges having an internal impedance close to 3 ohms (probably 2 to 5 ohms cartridges would work). It also means that the turns ratio of that cartridge is such that when mated to that group of cartridges will result in an acceptable ratio between the internal R of the cartridge and the input impedance that will be seen by the cartridge when using that SUT hooked up to the standard MM stage with a 47K ohm input impedance.
The turns ratio is equal to the voltage step-up ratio. A SUT with a turns ratio of 1:10 will increase the output signal V of the cartridge by 10-fold. Etc. The resulting impedance seen by the cartridge will be equal to the input Z of the phono stage (47K ohms) divided by the square of the turns ratio (100, in this case). So the cartridge sees 470 ohms (47,000/100). Since you want the ratio of the internal R of the cartridge to be about 1/10 of the input impedance of the phono, you can see that any cartridge with an internal R of less than 47 can drive a 1:10 SUT. Nothing explodes if you break the rule of thumb. The actual loss of gain and high frequency response that occurs when the match gives you a ratio less than 1:10 is on a continuum, losses get gradually worse as the ratio approaches 1:1. At a ratio of 1:1, 50% of cartridge output goes to ground instead of driving the phono stage, and there is audible loss of HF response, due to input capacitance.
Back to our "3 ohm" SUT. A theoretical cartridge with an internal R of 3 ohms can drive a net 30 ohm load before you cross the line of the 1/10 rule of thumb. In order to create a 30 ohm load, you just have to work the formula for the load seen by the cartridge backward. I calculate 1:40. A cartridge with a 3 ohm internal R can drive a SUT with a step-ratio (voltage gain) of 1:40 into a 47K load, if you want to rigidly obey the 1/10 rule of thumb.
By the way also, an ideal SUT or any transformer does not have an impedance per se. Transformers merely reflect the impedances from secondaries to primary windings, or vice-versa. I wondered that Elliot mentioned there are SUTs that let you choose both voltage gain and impedance. That can only be done if the SUT has built in resistors typically across the secondary, which would be in parallel with the standard 47K load.
Whart, you wrote, "My dumbed down approach is to look at the internal impedance of the cartridge and multiply by 10 for the load seen by the cartridge. The secondary (or output) from the SUT will be 47k." In your first sentence, I suppose you mean that you multiply by 10 to determine the (minimum) load that should be seen by the cartridge. I agree. In your second sentence, you may be confusing others. The secondary is connected to a 47K standard load at the input of an MM stage. The SUT acts on that impedance to reduce it as seen by the cartridge, in proportion to the square of the turns ratio of the cartridge. I am guessing you know this but there is a semantic disconnect in what you wrote. I point it out only so others will not be confused.