AFAIK the point about an isolation transformer is that there is no other connection of the secondary circuit to ground. If the user touches the electric supply (by fault or by error) it does not make a circuit through which current can flow and do him harm.
View attachment 234102
The secondary of the transformer is at an undefined potential wrt the ground and, when he touches part of that circuit, that point becomes grounded through him. A small transient current could flow as contact is made, but that would be to discharge any static potential that had arisen on the small capacitance of the wires. Once this is discharged, conduction through him maintains that point at ground potential. No significant current can flow, because there is no low resistance path to ground from anywhere else in the secondary circuit.
To get a dangerous shock he would need to touch two parts of the circuit, say either side of the tool, which were at 120 V pd.
View attachment 234103 here I measure the anode voltage (= 402.7 V wrt chassis 0 V ) by measuring the pd wrt to the +500 V line. I may do this because the anode resistor is simply a convenient place to attach my meter, or maybe my 100 V range (or other low range) is easier to read accurately than the 500 V range.
(BTW I can't think of any reason this circuit would be biased like this. Its just for the sake of argument ! Though perhaps I could make it into a class B/C amplifier with a few mods?)
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Edit: scrub the decimal of the voltages. Even on the 100 V range I couldn't read 0.3 V ! That was just me trying to show an accurate reading. I'd probably be able to estimate to 1 or 2 V on the 100 V range, but only to 5 or 10 V on the 500 V range (or the 600 V range which I would need to use on my meter.)