SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO

Answer: no. Each table corresponds to a pair of permutations $(s,t)\in S_n\times S_n$ such that $$s=\left(\begin{array}{cccc}1& 2& \dots & n\\ a_1& a_2& \dots & a_n\end{array}\right)\text{and}t=\left(\begin{array}{cccc}1& 2& \dots & n\\ b_1& b_2& \dots & b_n\end{array}\right).$$ The operation $A$ transforms a pair $(s,t)$ into the pair $(st,t)$, where the composition of the permutations $st$ is read from left to right – first the permutation $s$ is applied, and then the permutation $t$. Accordingly, the operation $B$ transforms a pair $(s,t)$ into the pair $(s,ts)$. For each pair of permutations $(s,t)$ we can consider the expression $st{s}^{-1}{t}^{-1}$, which is called the commutator of this pair and is denoted by $[s,t]$. Let us calculate the commutators of the pairs $(st,t)$ and $(s,ts)$: $$[st,t]=(st)t(st{)}^{-1}{t}^{-1}=stt{t}^{-1}{s}^{-1}{t}^{-1}=[s,t]\text{and}[s,ts]=s(ts)s^{-1}(ts{)}^{-1}=sts{s}^{-1}{s}^{-1}{t}^{-1}=[s,t].$$ Thus, the commutator is invariant under the operations A and B.

The table $$\begin{array}{ccccccc}2& 1& 3& 4& \dots & n-1& n\\ 2& 3& 4& 5& \dots & n& 1\end{array}$$ corresponds to a pair of permutations $(12)$ and $(12\dots n)$, which are cycles, and the table $$\begin{array}{ccccccc}2& 3& 1& 4& \dots & n-1& n\\ 2& 3& 4& 5& \dots & n& 1\end{array}$$ corresponds to a pair of permutations $(123)$ and $(12\dots n)$, which are also cycles. Their commutators are easy to find: $$[(12),(12\dots n)]=(12)(12\dots n)(21)(n\dots 21)=(12n),$$ $$[(123),(12\dots n)]=(123)(12\dots n)(321)(n\dots 21)=(123n).$$ Since they are not equal, Dima will not be able to get the second table from the first one.