Now, since V2 is say, twice V1 we get:

V1 + 2(V2) = V1' +2(V2)'

Umm... don't you mean:

V_1 + (-2 * V_1) = V_1' + V_2' (V_2' is *not* twice V_1, and V_2 is in the *opposite* direction of V_1)

-1 * V_1 = V_1' + V_2'

Since the two cars are completely inelastic and stick together, V1' and V2' are the same, so we can combine to get:

V1 + 2(V2) = 3V'

Which would make this:

-1 * V_1 = 2 * V' where V' = V_1' = V_2'

or:

V' = -0.5 * V_1

So the velocity of the new object (the two cars stuck together) is half the velocity of the slower car, heading in direction of the faster car, but I don't think that has anything to do with which of the two sustains the most damage. In that, I think that tanstaafl is correct. Imagine that instead of two cars, you have two non-identical objects that otherwise fulfill the parameters of the question. Object one is a 1-ton block of jello, object two is a one ton block of granite (suppose that the surface area where they collide is identical). Obviously, the one ton block of jello is going to sustain more damage than the one ton block of granite is, no matter which of the two objects is travelling faster. Now if you have two objects that have identical properties for dispersing the forces of impact, then they will absorb an identical amount of the forces of impact, due to that whole issue of relativity.