The electrode and the H2 is not two seperate entities in the transmission spectrum and the density of states.
The result will change as you increase the voltages, as you enforce a strong new external potential into the system.
Therefore the electrons will rearange according to new external field, this is the main idea behind ATK - that the system is allowed
to response to the applied voltage, where as many other codes only takes into account a shifting of the chemical potential.
This is the reason for ATK giving better IV curves than most other programs.
When looking at your system, it has some flaws in its setup, as they will most certain lead to incorrect results.
The electrode only has 2 atoms in it:
....
# -----------------------------------------------------------------------------
# Electrodes 0 and 1
# -----------------------------------------------------------------------------
# Index Element x (Ang) y (Ang) z (Ang)
0 Li 4.80 4.80 0.00
1 Li 4.80 4.80 2.88
...
This will give incorrect results even at 0 bias - it needs to be atleast 4 atoms longer, and I would personally go with 6 to be on the safe side.
The other problem is the number of atoms in the central region, it is too short in more than one sense. First of all, even in zero bias, you need to have enough Li atoms, before the H2 to make sure that the pertubation of the wire ( caused by the H2 ) is screened away at the boundary of the electrode. And as a 1d metal is the worst screening material in the world, you will need to have at least 6 atoms on both side of the H2 in order to get good transmission curves etc. But you only have 2...
# -----------------------------------------------------------------------------
# Index Element x (Ang) y (Ang) z (Ang)
0 Li 4.80 4.80 5.76
1 Li 4.80 4.80 8.64
2 H 4.80 4.80 11.80
3 H 4.80 4.80 10.82
4 Li 4.80 4.80 13.98
5 Li 4.80 4.80 16.86
When you apply the bias to the system ( and especially a large on of 6 V ) you will create an external field on the system, however the field strength of this external potential depends strongly on the length of the scattering region, in your case it is rougly 13-14 Ång long, giving a field strength that is in order of 0.5 V per Ångstrom ( 0.42*10^10 V/m ) which is an very strong field to introduce into the worst screening material in the world
In the real world, physical world, the system would disappear in a lighting strike, if one had success in enforcing this field onto a wire of this dimensions.
To solve the problem in this specific calculation, you will need alot to extended the dimension of the wire long enough to decrease the field strength to a more less wild field strength. I personally belive that a field strengt in a Li wire can not larger than 0.02-0.04 Volt/Ang before it becomes unstable to appling the current, therefore if I would to continue your calculation, I would increase the length of the scattering region by a factor 10-15 times by adding 20 Li atoms on each side of the H2.
