It depends on the design / implementation.
If the full power supply current that drives the light is switched through the remote switch, there will be an "I squared R" loss based on the total circuit resistance presented by the remote switch, its connector(s), and most obviously, the DC resistance of the cabling connecting the switch to the light, and proportional to the total current being drawn by the light. When operating, this will result in a "voltage drop" across that distributed resistance which reduces the supply voltage the light has available to operate on. How the light reacts to / deals with that reduced supply voltage is the question.
Will the internal circuitry of the light in question be able to completely compensate for that decrease in supply voltage such that it has no effect on the optical output of the light?
I don't know. Someone who's familiar with internal flashlight circuitry design can answer that.
Unlike that example, if the design is such that the full power supply current that drives the light is NOT going through / being switched by the remote switch, but instead a low level 'control' signal provided by the light is being switched, which the light's internal circuitry then uses to control the output of the light, then the distributed resistance of that switch, cabling, connection(s), etc. is not likely to have any impact on the output of the light at all.
EDIT: One case where the impact of a remote switch on a flashlight's output was discussed:
"As of August 2020, Surefire customer service says that the SR07 ST07 switch may not allow the M600DF to run at full brightness when using an 18650 cell." When I saw this statement, I decided to do a small test. Using my light meter, check how much the brightness will drop when using the tape...
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