We will show that we can use finite steps of the given transformation to transform $n$-tuple $(1,2,\dotsc ,n)$ to any cyclic permutation of $(1,2,...,n)$. We have \begin{align*} (1,2,3,4,5,6,\dotsc ,n) & \to (1,1,4,4,5,6,\dotsc ,n)\\ & \to (1,1,2,6,5,6,\dotsc ,n) \\ & \to (1,1,2,3,8,6,7,\dotsc ,n) \\ & \vdots \\ & \to (1,1,2,3,\dotsc ,n-2,2n-2) \rightarrow (n,1,2,3,\dotsc ,n-2,n-1). \end{align*}From here we can create any cyclic permutation of $(1,2,\dotsc ,n)$. Back to the problem. Starting with $(1,2,\dotsc ,2015)$, we transform it to a cyclic permutation that $2015$ is in the place we want at the end. Then we use the same argument ignoring $2015$ which is already in place and consider only $(1,2,\dotsc ,2014)$ which is now in cyclic permutation, and so on.

ThE-dArK-lOrD wrote: We will show that we can use finite steps of the given transformation to transform $n$-tuple $(1,2,\dotsc ,n)$ to any cyclic permutation of $(1,2,...,n)$. We have \begin{align*} (1,2,3,4,5,6,\dotsc ,n) & \to (1,1,4,4,5,6,\dotsc ,n)\\ & \to (1,1,2,6,5,6,\dotsc ,n) \\ & \to (1,1,2,3,8,6,7,\dotsc ,n) \\ & \vdots \\ & \to (1,1,2,3,\dotsc ,n-2,2n-2) \rightarrow (n,1,2,3,\dotsc ,n-2,n-1). \end{align*}From here we can thus create any cyclic permutation of $(1,2,\dotsc ,n)$. Back to the problem. Start with $(1,2,\dotsc ,2015)$, we transform it to a cyclic permutation that $2015$ is in the place we want at the end. Then we use same argument ignoring $2015$ which is already in place and consider only $(1,2,\dotsc ,2014)$ which is now in cyclic permutation and so on. Very nice solution dark lord!!! I have another solution. We will make induction. n=2 clearly true. Assume that for n=k-1 true and we prove for n=k. İf we can relocate i and k (1≤i≤k-1) it's not hard to see that we are done. İf k=2m we take i=m İf k=2m+1 we take $i=2^t +1-k$ t:$2^t-1<k<2^t$ We use modulo $2^t +1$.Take an arbitrary integer x. $(x,2^t +1-x)\implies(x+\frac{2^t +1-x}{2},\frac{2^t +1-x}{2}) or (x,2^t +1-x)\implies(\frac{x}{2},2^t +1-x+\frac{x}{2})$ anyway $ x\implies\frac{x}{2}(mod2^t +1)$ So if we apply our process t time to pair of (k,i) clearly we obtain pair of (i,k) and we are done

ThE-dArK-lOrD wrote: We will show that we can use finite steps of the given transformation to transform $n$-tuple $(1,2,\dotsc ,n)$ to any cyclic permutation of $(1,2,...,n)$. We have \begin{align*} (1,2,3,4,5,6,\dotsc ,n) & \to (1,1,4,4,5,6,\dotsc ,n)\\ & \to (1,1,2,6,5,6,\dotsc ,n) \\ & \to (1,1,2,3,8,6,7,\dotsc ,n) \\ & \vdots \\ & \to (1,1,2,3,\dotsc ,n-2,2n-2) \rightarrow (n,1,2,3,\dotsc ,n-2,n-1). \end{align*}From here we can create any cyclic permutation of $(1,2,\dotsc ,n)$. Back to the problem. Starting with $(1,2,\dotsc ,2015)$, we transform it to a cyclic permutation that $2015$ is in the place we want at the end. Then we use the same argument ignoring $2015$ which is already in place and consider only $(1,2,\dotsc ,2014)$ which is now in cyclic permutation, and so on. Why can we create any permutation of (1, 2, 3, ... n)?