Hi!
"... a few diodes in series..." I'm not sure why you think you would need a few diodes?
You would need one diode in series.
A diode in series will sink your voltage down to probably 0.7v that would mean around 4.3v from 5v.
... not at all.
Contrary to the common misbelief, a diode is not a constant voltage drop. It's a logarithmic function of the current. A digital camera is expected to draw as little as dozens of microamperes when shut down, to several amperes peak (for example, when storing to the memory card and charging the xenon flash capacitor simultaneously). Given a "typical" silicon diode that can handle the peak current, the voltage drop across the diode would vary between around 0.4V and 0.9V, typically. If the camera performs an overvoltage self-check
before having significant current draw, this could be a problem.
Worse, add any temperature variations in the play, and the range gets wider: higher temperature lowers the drop, lower temperature increases the drop. In the end, while it's somewhat common to use diodes in this class of situations ("almost the right voltage; need to drop a bit"), often it doesn't work out when you do the actual numbers over the full actual operating conditions and realize your "0.7V" suddenly becomes something between 0.3 and 1V.
In this case, the diodes
could perhaps work if you should expect the full working voltage range of the li-ion cell to work; this means
at least from 3.3V to 4.1V (probably from around 3.0V to 4.2V). You just need to aim lower than 4.1V nominal. Assuming a diode drop is min 0.4V, max 0.8V, and assuming the 5V is actually between 4.9V and 5.1V, using two diodes would vary from 3.3V to 4.3V. This already uses slightly optimistic assumptions and still results in illegal output range (to 4.3V), so it has negative margin for error, but yes, two diodes in series
could barely work by luck. One is definitely not going to (or if it does, then the whole premise is meaningless; it's not sensitive to the voltage at all and would probably work with 5V directly as well!)
The 3.6V linear regulator suggestion is a good one; just use a beefy one, I
guess something specified to 1.5A nominal (therefore current limit way over 2A) could work. The peak current can be more than you think, and the peaks are long enough (~seconds) that a large electrolytic capacitor on the regulator output is not the answer.