I agree with the others that have pointed out the tremendous power requirement for heating water for a shower, suggesting that something doesn't quite add up to make one shower work and one not. 3600W simply will not heat up much water flow by very many degrees.
In addition to checking the actual voltage and current at the heater that isn't working, I'd look for a wiring error at the one that is working...perhaps it was wired 208V instead of the rated 120V, and with water flow might actually function for a while despite the overload involved.
Since Rob asked about sizing a buck-boost, I'll focus on answering that.
1) While you boost the supply voltage to compensate for voltage drop, this is generally considered a bad idea because the boost is present even when the load is off and the voltage drop goes away. You have to make sure that nothing is damaged by excessive voltage when the load is off.
2) Since transformers work by trading voltage for current, you want to place the transformer _before_ your long wire run. If you put the transformer near the load, then the increased current flow would mean increased voltage drop in the wires, requiring more boost.
3) Since the load is resistive and voltage drop is caused by resistance int he wires, the easiest way to calculate the necessary voltage is to figure out the voltage needed to push the rated current through the series resistance of the wire and the load. For simplicity I approximate the resistance of the wire and the load as constant.
Per Nec Table 8, 10 AWG wire has a resistance of 1.21 ohms per 1000 feet. So your circuit wires have a resistance of 0.484 ohms.
The rated current is 30A, and R=E/I R=120/30 = 4 ohms.
Your total circuit resistance is 4.484 ohms.
You want a current of 30A in this circuit, so you need a supply voltage of 134.5V, and a supply power of 4036W.
(Note: with 120V supply, I get 26.8A through the circuit, 107V at the water heater, and 2860W delivered)
4) The 'trick' to buck-boost transformers (noted post 5) is that you size the transformer for the boosted part of the load, not the whole load. The 'unboosted' portion of the load is supplied directly, bypassing the transformer.
You have a 120V supply, and you need to add 14.5V to it, with 30A of capacity. Unless you get something custom wound, in practice you will get a transformer with a 12V or 16V secondary; pick your poison, slightly too high or slightly too low. This transformer would need about a 480VA rating (secondary current * secondary voltage).
Something like
https://www.zoro.com/square-d-transformer-120240vac-1632vac-500va-500sv46b/i/G0734361/ would be suitable.
5) The supply up to the transformer will draw about 34A.
Given that you are looking at about $150 for a transformer to feed a $150 water heater, you might consider replacing the water heater with a similar 240V version. The % voltage drop will be halved, and the rated power will be doubled. With the Chronomite-SR-30-240, and a 240V supply, you would get 226V at the heater and 6400W delivered.
-Jon
In this case installing the buck/boost at the panel end would be much easier.
