Let $k$ be a positive integer. Let $A=\{a_1,\dots, a_{2^k}\}$ be a subset of $\mathbf{Z}/2^{k+1}\mathbf{Z}$ whose image in $\mathbf{Z}/2^k\mathbf{Z}$ is the whole $\mathbf{Z}/2^k\mathbf{Z}$. Let $B=\{b_1,\dots, b_{2^k}\}$ be a subset with the same property. Is it always possible to find a permutation $\sigma$ on $2^k$ letters such that $$a_1+b_{\sigma(1)},\dots, a_{2^k}+b_{\sigma(2^k)}$$ are pairwise distinct?
A little background. (A weaker version of) Snevily's conjecture asserts that, if $l<n$ and $n$ is odd, then for any two subsets $A=\{a_1,\dots,a_l\}$, $B=\{b_1,\dots, b_l\}$ of $\mathbf{Z}/n\mathbf{Z}$, there is always a permutation $\sigma$ on $l$ letters such that $$a_1+b_{\sigma(1)},\dots, a_{l}+b_{\sigma(l)}$$ are pairwise distinct. This has been proved in full generality by Arsovski. The theorem is certainly false if $n$ is allowed to be even. Intuitively, it's the power of $2$ in $n$ that will cause the problem. So I only look at $\mathbf{Z}/2^k\mathbf{Z}$ and want to see what can be said. The above question is my gut feeling.