That would be a sensible inference, yes, if utilities were elements of a vector space. This is exactly why I said the terminology "utils" was so misleading -- it suggests that they work like meters, seconds, etc; that we can think in terms of units and dimensions. But that doesn't work here!

Dimensional analysis is basically analysis of scaling symmetries. It works when the only symmetries are scaling symmetries. But utilities, being an affine thing rather than a linear thing, have more symmetries than that! They have translation symmetries too! Units and dimensions are a very useful tool but they are not universal and they can't handle something like this, unless you make sure you carefully distinguish between utilities and utility differences.

(That last one seems to be the usual solution to this sort of problem -- not in the case of utitilities but more generally. E.g. we don't hesitate to talk about points in time as quantities of seconds, even though time is translation-invariant, but really it's only durations that are quantities of seconds. When we describe a point in time as being at "5 seconds", we really mean "5 seconds after some agreed upon starting point". And so while adding or halving durations is meaningful, adding or halving positions in time is not, because what's going on with the starting point? (By contrast, averaging positions in time is meaningful, as is 2*t_1 - t_2, etc.) But time is familiar, so people don't tend to make that sort of mistake, of forgetting that times are measured relative to some implicit arbitrary baseline; whereas utility is not so familiar, and, well, you're making that mistake right now.)

If you like, you can imagine -- as I've essentially done in my post above -- that this affine space is embedded in some larger vector space, like the line x+y=1 embedded in R^2, and that elements of x+y=k have type "k*utility".

But this is becoming stupid. This is a hell of a lot of words; the fact of the matter is that if you were right, then we could take two outcomes a and b, with u(a)=1, u(b)=2, and observe that 3u(a) > u(b); then define an equivalent utility function v by v(x)=u(x)-2, and observe that now 3v(a) < v(b), so apparently in fact v was not equivalent. I.e. if you were right, then utility functions would only be unique up to positive scaling, not up to general positive affine transformations.

Only one of the following can be true: 1) It is meaningful to take non-affine combinations of utility functions 2) Two utility functions related by a positive affine transformation are equivalent

And it's the latter. Why? Because if you look at the definition of what it means for a function u to be the utility function of a given agent, you'll notice it only involves comparisons of affine combinations of values of u, not general linear combinations. Hence, applying any positive affine transformation will not change the comparisions, and the result will again be a utility function for the given agent.

And this is why everyone says that they're only unique up to positive affine transformation, and correct in saying so. If the definition of a utility function relied on more general linear combinations of utilities, then that would restrict the symmetries further, and it would probably result in there only being scaling symmetries, in which case you would be right.