How to switch between 2 power supplies?
Switching between two supplies is more complicated than it sounds on the first moment: You need to do two things at the same time: You need to turn off the connection to one suppy and turn on the connection to the other supply. If you turn off too early, power supply to the sensor gets interrupted for a short moment, while if you turn on too early, you are forcing 5V into the 2V supply.

Aftter providing your suggested circuit, I strongly recommend a different approach: Build one(!) supply you can switch between 5V and 2V output. If you are fine with linear regulators and have enough headroom, this might be an LM317 with a split resistor in ground path (either two parallel resistors where one gets interrupted by a logic level N-MOSFET or two in series where the lower one gets shorted by such a MOSFET).

You get the 2V supply from a MAX603 by using a resistor divider: This shows you how to control the resistor devider from the arduino to change the output voltage on-the-fly. 1M/330k (output high, FET conducting) results in 4.83 volts. 1M/1530M (output low, FET not conducting) results in 1.98V. By using custom resistors (or paralleling 2 of them) instead of single off-the-shelf E6 values, the targets of 5V/2V can be obtained more acurately.







simulate this circuit &ndash; Schematic created using CircuitLab

You can actually use any kind of logic level N-channel MOSFET, i.e. a FET that is guaranteed to have a on-state resistance that is negigible at a gate voltage obtainable from the arduino. The IRF730 is not really logic-level, as its performance (see the output characteristic diagram in the ST datasheet) is sub-par at 4V. Most likely it is that much oversized that it still provides way enough conductance even at 4.5V (and you should get that from a 5V-powered uC). The IRF540 is similar in behaviour. The 2N7000 in that circuit is slightly better at specifications (they specify at least 1mA current at 3V gate voltage, while the IRF730 and IRF540 specify at least 0.25mA current at 4V), but also not really a logic-level FET. In your case, the current is 1.2V across a 1M resistor, so it's 0.0012mA, so at least a factor 200 below the currents mentioned the data sheets, so conductance should suffice anyway.

I still recommend getting a bag of 2N7000 anyway, as these parts only cost some cents. They really are static sensitive though, so handle them with care and have some spare ones at hand if you use them and regularly plug them by hand.