Find an explicit bijection between the spaces of sequences of $\{0, 1, 2, 3, 4, 5\}$ of sequences of $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$. We know there is a bijection between these two space. We even know there is a topological conjugacy if we consider adding machines on them. But I would like to see an explicit bijection between the space of sequences of the set $\{0, 1, 2, 3, 4, 5\}$ and the space of sequences of the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$.
If we have, for example, spaces of sequences of sets $\{0, 1\}$ and $\{0, 1, 2, 3\}$ it is easy to find a bijection between these two spaces. Simply code numbers as follows:
$$00 \mapsto 0, 10 \mapsto 1, 01 \mapsto 2, 11 \mapsto 3.$$
Hence for example the sequence $(001100110011\ldots) \mapsto (030303\ldots)$.
It is always easy to find this coding if we have sets of the forms $\{0, 1, \ldots, j - 1\}$ and $\{0, 1, \ldots, j^n - 1\}$. But I don't know how to find it in more general cases.
Edit: being equivalent means there is a bijection and it is important to explicitly find a bijection.
 A: We can consider sequences of $\{0,1,\dots,5\}$ and $\{0,1,\dots,11\}$ as the digits in base 6 and 12 of real numbers in the interval $[0,1]$. So we can try taking a sequence of $\{0,\dots,6\}$, turn it into a real number, then take the base 12 expansion of this real number to get a sequence of $\{0,1\dots,11\}$. This almost gives a bijection, but some numbers will have two expansions (for instance, in base 6 $.05555\dots=.10000\dots$). These overlapping cases happen exactly when one of the expansions has only finitely many non-zero digits, so we can deal with those cases separately. If we have finitely many non-zero terms, we can take the last non-zero term, and read backwards from there to get a unique finite sequence. Now as Tom mentioned in the comments, we do have a straight forward bijection between finite sequences, just treat them as expressions of integers in base 6 and 12.
So, to summarize, our bijection takes some sequence of $\{0,1,\dots,5\}$, if it has infinitely many non-zero terms it takes this to the real number you get by treating the sequence as the digits in base 6, then takes this real number to the sequence you get from its digits in base 12 (choosing the expansion with infinitely many non-zero terms if there is more than one base 12 expansion). If it has finitely many non-zero terms, it starts from the last non-zero term and reads backwards from there to get a number in base 6, then converts this to base 12 and takes it to the sequence you get from writing out the digits backwards.
A: Speculations too long for a comment.
I agree that the accepted answer from
@AlexanderTenenbaum works. However, deciding whether a sequence contains only a finite number of $0$s requires examining the whole thing - so no sequential algorithm can construct the image of a sequence under this bijection.
Contrast that with the OP's example. Whenever the symbol sets have cardinality $j$ and $j^n$ there's a bijection between lists that is constructed sequentially.
I wonder whether that's the only case where such an algorithm exists. How would you formulate that conjecture? This touches on the old philosophical arguments on completed versus potential infinities.
