How much more will the cost be for a wager on V5 compared to V6 when adjusting for difference in probability, price per unit and track's deduction?

This sounds very complex. You aren't going to alter the probabilities of the results by betting any more or less money on a race.
You will obviously alter your expected payout 'if you do win'.
Are you trying to achieve the same payout from each bet so you are looking for how much to stake in each case?

Like if you bet 3 units at 4:1 you will need to bet 6 units at 2:1 to get the same payout?
You can add a bit of complexity to this simple example by saying the bookie takes a different percentage for each bet and the numbers won't be quite as simple as 3 and 6.
This seems to be the question you are asking but I can't be certain.

Your assumption is to some extent what I mean. My intentions are:

1. Establish how many more units are required to achieve the same overall probability as another type of wagering (for example, how many more units are required for V75 to achieve the same overall probability as V65/V64?).

2. Establish the cost for doing 1 above with consideration taken to cost per unit and the track's deduction.  

For example:

V4:         cost 2 SEK/unit               track's deduction -25%                    overall probability 1/20736 (12x12x12x12)            
V5:         cost 1 SEK/unit              track's deduction -35%                    overall probability 1/248832 (12x12x12x12x12)            
V64:       cost 1 SEK/unit              track's deduction -35%                    overall probability 1/2985984 (12x12x12x12x12x12)            
V65:       cost 1 SEK/unit              track's deduction -35%                    overall probability 1/2985984 (12x12x12x12x12x12)            
V75:       cost 0.50 SEK/unit              track's deduction -35%                    overall probability 1/35831808 (12x12x12x12x12x12x12)      

Actually, it's the same bookie (they have monopoly on wagering, ATG), but they deduct differently depending on the type of wagering.

This is what I have been able to conclude so far:

V75: I need to spend 12 times more units compared to V65 and V64 in order to achieve the same overall probability. But the cost to achieve this is only 6 times more, because the cost per unit for V75 is 0.50 SEK whereas the cost per unit for V65 and V64 is 1 SEK per unit. No need here to calculate difference in track's deduction as it is 35 % for both types of wagering.

V65 and V64: I need to spend 12 times more units compared to V5 in order to achieve the same overall probability. The cost to achieve this is also 12 times more, because the cost per unit for V65 and V64 compared to V5 is 1 SEK for all three types of wagering (the same). And also here is no need to calculate difference in track's deduction as it is 35 % for all three types of wagering (the same).

V5: I need to spend 12 times more units compared to V4 in order to achieve the same overall probability. Here, I am not sure how to calculate the cost to achieve this:

First, it's the same situation as with V75 in that the cost is only 0.5 of the cost for V4, which means the cost would be only 6 times more. But then, how do I account for the fact that the track's deduction is 25 % for V4 and 35 % for V5? Am I correct if I say the cost will be 6.4 times more for V5 compared to V4 to achieve the same overall probability?

I'm OK with the idea of a slightly complex equation, one that returns a fixed amount when taking into account the different odds and the variations in track deductions.
What I can't quite follow is where you say "to achieve the same overall probability", you can't change the probability of the results. What you can change is your expected return,  for instance if you covered every possible outcome to guarantee a winning bet.

What I mean with "to achieve the same overall probability" is this:

                              Achieve same probability:                                 Achieve same probability:                        
V4:   2 SEK/unit                                                                                                                              track's deduction -25%   overall probability 1/20736            
V5:   1 SEK/unit    requires 12 times more units than V4             costs 6.4 times more     track's deduction -35%   overall probability 1/248832          
V64: 1 SEK/unit    requires 12 times more units than V5             costs 12 times more      track's deduction -35%   overall probability 1/2985984            
V65: 1 SEK/unit    requires 12 times more units than V5             costs 12 times more      track's deduction -35%   overall probability 1/2985984            
V75: 0.5 SEK/unit requires 12 times more units than V65/V64   costs 6 times more        track's deduction -35%   overall probability 1/35831808  

 For example, the probability to "achieve" success for V75 is 1/35831808 whereas the probability to "achieve" success for V65 is 1/2985984, and the difference is 1/35831808 divided with 1/2985984, which is 12 times, thus 12 times more units are required for V75 to "achieve" success at the same rate as for V65. But because the cost per unit differs (0.5 SEK for V75; 1 SEK for V65), the cost is only 6 times more for V75. Track's deduction is the same for V75 and V65 (but between V5 and V4, this differs, as between other combinations with V4, for example V75 and V4 and so on).

Thibault, perhaps we are talking about the same thing but using other words; you say expected return, I say overall probability. This is all objective, not subjective.

The assumption is that there are always 12 horses in each field (which actually could add a fourth variable, based on statistics; for instance to assume x number of horses in each field for the wagering form V5 based on statistcs. and to assume y number of horses in each field for the wagering form V4; this is very relevant for the wagering form V6 for example, where the fields often are very small, consisting often of only 8 or even less horses).

Anyway, for this question the assumption is that there are always 12 horses in each field for any of the different wagering forms. That gives on objective overall probability for winning on V4 of 1/20736; on V5 1/248832, etc. Are you following me now?