Electronic – Connecting multiple DC sources in parallel without the use of diodes

Your results are inconsistent with expected operation.
Either you are not doing what you think you are doing or the MOSFET is damaged or your test meter is of very low "ohms per volt".

Test1: Connect test meter with -ve probe to ground and +ve probe via 1 k to 3V3.
What is the voltage reading?

This should read 3V3 to a very close approximation.
If it does not give or throw the meter away and get a slightly better one :-).

Any meter that reads wrongly in that situation is a VERY poor one and useful only for eg battery testing.

Test2: Set meter to diode test range.
Measure Drain - Source.
With Source = +ve you should see a diode with Vf higher than a usual silicon diode.
With +ve on Drain you should see O/C.
With metetr connected either way G-D and G-S you should get open circuit.

Test 3 Ask Olin for advice.

Test 4: Check your circuit carefully.
Recheck MOSFET pinouts.

Try a new FET.

Note that MOSFETS are VERY prone to ESD damage - especially gate to D or S.
Handle with proper electrostatic precautions.

Report back.

ICMRmax is set by keeping M5 in saturation. That implies that the maximum voltage at the drain of M5 is \$VDD - V_{dsSatM5}\$. M2 and M3 are PMOS, so their source will be their most positive terminal and their gate voltage must be less than their source voltage if they are to operate properly. That means:

$$ ICMRmax = VDD - V_{dsSatM5} - V_{gsM2} $$

\$V_{gsM2}\$ is set by the current flowing through M2. Your question implies that the smallest possible value of \$V_{gsM2}\$ is \$V_{t}\$. Strictly, that's true, but since your bias current is already defined and is not zero, \$V_{gsM2}\$ will be larger than \$V_{t}\$. However, for the moment, assuming it is \$V_{t}\$, then ICMRmax = 1.8 - 0.5 - 0.5 = 1V (I assume from your equation that VDD is 1.8V)

ICMRmin is defined by keeping the branch with M2 and M4 in saturation. For this to be true, minimum voltage at the drain of M4 is \$V_{dsSatM4}\$. For M2 to be at the point of saturation

$$ V_{dsSatM2} = V_{gsM2} - V_{tM2} $$

ICMRmin will be the gate voltage of M2 under these conditions. This is given by:

$$ ICMRmin = V_{dsSatM4} + V_{dsSatM2} - V_{gsM2} $$ $$ = V_{dsSatM4} - V_{tM2} $$ $$ = 0.2 - 0.5 $$ $$ = -0.3V $$

That makes your ICMR -0.3V to 1V. Actually it will be narrower than that, because \$V_{gs}\$ will be larger than \$V_{t}\$ as mentioned.

I think that is correct! It's a bit counter-intuitive that the ICMRmin can be negative, but it's a common result for diff amps with PMOS input (you can find it in the textbooks or elsewhere on the web).