The sum of 1 and 12 ensures that there is at least one 0 and one 6. Since the sum ranges from 1 to 12 only, the faces of the modified die can only have an integer from 0 to 6. This also means that the sum of two outcomes can be 1 only iff the regular and modified dice outcomes are 1 and 0 respectively.

Let $R$ and $M$ be the outcomes of the regular and modified dice respectively.

$\implies P(R + M = 1) = P(R=1 \cap M = 0)$

$= P(R=1) \cdot P(M = 0)$

$\implies 1/12 = 1/6 \cdot P(M = 0)$

$\implies P(M = 0) = 1/2$

Similarly, considering $P(R+M=12) = 1/12 \implies P(M=6) = 1/2$

These two marginal probabilities sum to 1, i.e., there is no other number on any of the faces of the modified die. The modified dice must have three 0's and three 6's.