> I want to build a simple rotary switch attenuation box ... Wouldn't that give me the best voltage transfer? what am I overlooking?
If you want to attenuate, you do NOT want best voltage transfer. You want "bad" voltage transfer. Duh!
It is very possible your input will not be happy looking out at 12K. Most modern inputs want a low-Z source to damp-out crap on the line and in the input wiring. It is rare to find a modern output that is as high as 600 ohms actual-impedance.
As Jakob says: you want an attenuator that has an input impedance no lower than the rated load of the source (possibly 600Ω), and the lowest possible output impedance, and known voltage-ratios. The best approximation to this is a Potentiometer. Given those exact values, use a 631.57Ω pot. If you use switching instead of a rubby-pot, that's your total resistance. But since most outputs are just as happy with 1K||12K= 923Ω loading, a 1K pot is convenient. Or try a 5K pot: the output impedance can be as high as 1.25K, wich is still "low" compared to 12K loads. And significantly lower than the 6K line impedance you would get with a series resistor pad and -6dB attenuation.