I already have a more detailed version here; see the different calcualtions for E[T] vs E[IT]. However, I'll give you a short version. From the gnome's perspective, the two different types of total utilitarian utility functions are:
T = total $ over both cells
IT = total $ over both cells if there's a human in my cell, 0 otherwise.
and the possible outcomes are
p=1/4 for heads + no human in my cell
p=1/4 for heads + human in my cell
p=1/2 for tails + human in my cell.

As you can see, these two utility functions only differ when there is no human in the gnome's cell. Moreover, by the assumptions of the problem, the utility functions of the gnomes are symmetric, and their decisions are also. UDT proper doesn't apply to gnomes whose utility function is IT, because the function IT is different for each of the different gnomes, but the more general principle of linked decisions still applies due to the obvious symmetry between the gnomes' situations, despite the differences in utility functions. Thus we assume a linked decision where either gnome recommends buying a ticket for $x.

The utility calculations are therefore
E[T] = (1/4)(-x) + (1/4)(-x) + (1/2)2(1-x) = 1-(3/2)x (breakeven at 2/3)
E[IT] = (1/4)(0) + (1/4)(-x) + (1/2)2(1-x) = 1-(5/4)x (breakeven at 4/5)

Thus gnomes who are indifferent when no human is present (U = IT) should precommit to a value of x=4/5, while gnomes who still care about the total $ when no human is present (U = T) should precommit to a value of x=2/3.

Note also that this is invariant under the choice of which constant value we use to represent indifference. For some constant C, the correct calculation would actually be
E[IT | buy at $x] = (1/4)(C) + (1/4)(-x) + (1/2)2(1-x) = (1/4)C + 1-(5/4)x
E[IT | don't buy] = (1/4)(C) + (1/4)(0) + (1/2)(0) = (1/4)C
and so the breakeven point remains at x = 4/5