min() never does rounding. min() of a single parameter uses bit-for-bit comparisons (since the entire array is always the same datatype.)
== between two values of the same datatype always uses bit-for-bit comparisons. The output of min() over a single numeric parameter is always going to be the same datatype as the parameter was, and then you are comparing that bit-for-bit identical copy of the minima that is stored in minValue against the values in the original array. Except in the case where the array is all nan, the == in that situation is guaranteed to match at least one value. (When the entire array is nan, the output of min() is not guaranteed to be bitwise identical to any of the input nan values.)
Therefore, the problem is not in findAllMin()
However, the challenge could be in the values that are passed to findAllMin(). Different processors have different instruction sets and different cache behaviors, so it is expected that the exact results of numeric computation might be different on different processors.
The 7820 appears to have "One or Two Intel Xeon Scalable Processor family CPUs (1st and 2nd generation)with up to 28 cores per processor". However, current online purchasing options for it include Intel Xeon Bronze 3204, Intel Xeon Silver 4208, Intel Xeon Silver 4214R, Intel Xeon Gold 5222
The 7910 has "Single or dual Intel® Xeon® processor E5-2600 v4" . Unfortunately the 7910 does not appear to be sold anymore so I cannot list specific CPUs.
The m6800 has "4th Generation Intel® Core™ i5 and i7 processor options up to Intel Core i7 Extreme Edition"
When I compare the Xenon processors, it looks to me as if there are differences in the number of cores, in the efficiency, and in the amount of cache available per core, but I do not notice any differences in available instruction sets (I might have overlooked something.) However, differences in the amount of cache can lead to differences in computed result, as the calculations can end up getting blocked differently.
At the moment I have no ideas as to why you might be getting nans; I am not familiar with the algorithms being used. But I do know that differences are to be expected. I would tend to only expect a difference to result in a nan if you were working with calculations that are numerically unstable.