A well known congruence of Wolstenholme states that $$\frac1{1^2}+\frac1{2^2}+\cdots+\frac1{(p-1)^2}\equiv0\pmod{p}$$ for any prime $p>3$. For each $n=3,4,\ldots$ we clearly have $$\frac1{1\times2}+\frac1{2\times3}+\cdots+\frac1{(n-1)n}+\frac1{n\times1} = 1.$$

Motivated by the above, here I ask a new question.

**Question.** Is it true that for each prime $p>3$ there is a permutation $\pi\in S_{p-1}$ with $\pi(p-1)=p-1$ and $\pi(p-2)=p-2$ such that the congruence
$$\frac1{\pi(1)\pi(2)}+\frac1{\pi(2)\pi(3)}+\cdots+\frac1{\pi(p-2)\pi(p-1)}+\frac1{\pi(p-1)\pi(1)}\equiv0\pmod{p^2}$$
holds?

For $p=5$, there is a unique permutation $\pi\in S_4$ meeting the requirement, namely, $$\frac 1{2\times1}+\frac1{1\times3}+\frac1{3\times4}+\frac1{4\times2}=\frac{25}{24}\equiv0\pmod{5^2}.$$ For $p=7$, there is also a unique permutation $\pi\in S_6$ meeting the requirement, namely, $$\frac1{2\times3}+\frac1{3\times4}+\frac1{4\times1}+\frac1{1\times5}+\frac1{5\times6}+\frac1{6\times2}=\frac{49}{60}\equiv0\pmod{7^2}.$$ For $p=11$ there are totally $323$ permutations $\pi\in S_{10}$ meeting the requirement. For $p=13$, the permutation $$(\pi(1),\ldots,\pi(12))=(1,2,3,7,4,9,5,8,10,6,11,12)$$ meets our purpose. Based on these data, I conjecture that the question has a positive answer.

Your comments are welcome!