Consider the finite analogue where we model $\mathbb{Z}/b^n \mathbb{Z}$ as the ring $R = \mathbb{Z}[T]/\langle T^n \rangle$ modulo the ideal generated by $1-bT$. Note that $1$ in $R$ corresponds to $b^{n-1}$. Reversing sequences, so we get the more usual representation, we take the abelian group $\mathbb{Z}^n$ and quotient by the subgroup generated by all

$$ (0,\ldots, 0, -b, 1, 0, \ldots, 0) .$$

A set of representatives for the cosets is the $b$-ary words of length $n$. These are in bijection with the leaves of the rooted $b$-ary tree with $n$ levels. The automorphism group of the tree is the iterated wreath product $S_b \wr \ldots \wr S_b$, with $n$ factors. Since addition respects congruence modulo powers of $b$, addition of a constant is an automorphism of the tree. This already gives some connection.

To bring in carries, consider the binary case. Now addition of $1$ is the automorphism which swaps the two halves of the tree, and then continues on the (new) right-hand subtree, by swapping its two halves, and then continues working further down the tree. (For instance, after the first swap, an even numbered leaf $2m$ is replaced with the odd numbered leaf $2m+1$, after the second, an odd numbered leaf $4m+1$ is replaced with $4m+2$, and so on.) I think this repeated swapping is essentially the carry relation.

To make this explicit take $b=2$ and $n=3$. Then the leaves of the tree are labelled $0,4,2,6,1,5,3,7$, corresponding to the binary words

$$000, 001, 010, 011, 100, 101, 110, 111.$$

Addition of $1$ is the permutation $(0,1)(4,5)(2,3)(6,7)$ followed by $(0,2)(4,6)$ followed by $(0,4)$. (Confusingly the elements in these permutation come from successive left sub-trees, because the labels change as we apply the automorphism.) Chasing $3$, we get $3 \mapsto 2 \mapsto 0 \mapsto 4$. This corresponds to

$$110 \mapsto 210 \mapsto 020 \mapsto 001$$

in the carries model. For $7$ the carries model requires four steps

$$111 \mapsto 211 \mapsto 021 \mapsto 002 \mapsto 000$$

rather than the three predicted by the permutations, but this should be expected because we truncated the elegant infinite construction in the paper at $n=3$.

Consider the finite analogue where we model $\mathbb{Z}/b^n \mathbb{Z}$ as the ring $R = \mathbb{Z}[T]/\langle T^n \rangle$ modulo the ideal generated by $1-bT$. Note that $1$ in $R$ corresponds to $b^{n-1}$. Reversing sequences, so we get the 'little-endian' representation, we take the abelian group $\mathbb{Z}^n$ and quotient by the subgroup generated by all

$$ (0,\ldots, 0, -b, 1, 0, \ldots, 0) .$$

A set of representatives for the cosets is the $b$-ary words of length $n$. These are in bijection with the leaves of the rooted $b$-ary tree with $n$ levels. The automorphism group of the tree is the iterated wreath product $S_b \wr \ldots \wr S_b$, with $n$ factors. Since addition respects congruence modulo powers of $b$, addition of a constant is an automorphism of the tree. This already gives some connection.

To bring in carries, consider the binary case. Now addition of $1$ is the automorphism which swaps the two halves of the tree, and then continues on the (new) right-hand subtree, by swapping its two halves, and then continues working further down the tree. (For instance, after the first swap, an even numbered leaf $2m$ is replaced with the odd numbered leaf $2m+1$, after the second, an odd numbered leaf $4m+1$ is replaced with $4m+2$, and so on.) I think this repeated swapping is essentially the carry relation.

To make this explicit take $b=2$ and $n=3$. Then the leaves of the tree are labelled $0,4,2,6,1,5,3,7$, corresponding to the binary words

$$000, 001, 010, 011, 100, 101, 110, 111.$$

Addition of $1$ is the permutation $(0,1)(4,5)(2,3)(6,7)$ followed by $(0,2)(4,6)$ followed by $(0,4)$. (Confusingly the elements in these permutation come from successive left sub-trees, because the labels change as we apply the automorphism.) Chasing $3$, we get $3 \mapsto 2 \mapsto 0 \mapsto 4$. This corresponds to

$$110 \mapsto 210 \mapsto 020 \mapsto 001$$

in the carries model. For $7$ the carries model requires four steps

$$111 \mapsto 211 \mapsto 021 \mapsto 002 \mapsto 000$$

rather than the three predicted by the permutations, but this should be expected because we truncated the elegant infinite construction in the paper at $n=3$.