Let $n\geq 1$. Question : is it true that for any permutation $\sigma$ of $\frac{\mathbb Z}{2n{\mathbb Z}}$, there are two indices $i,j$ such that $1\leq i < j \leq 2n$ and
$$ \sigma(j)-j = \sigma(i)-i $$ I call such a pair $(i,j)$ good for $\sigma$.
My thoughts :
With the help of a computer, I have checked that the answer is yes for $n\leq 5$.
If $\sigma$ is the particular permutation defined by
$$ \sigma(k)=\begin{cases}2k-1,& \textrm{if }\ k\leq n, \\ 2(k-n)& \textrm{if }\ k >n.\end{cases} $$
Then there is only good pair, namely $(1,2n)$.
See Exercise 7 (a) in my Fall 2018 Math 5707 midterm #3 solutions (page 46). Here is the relevant part:
(Here, I am using the notation $\left[n\right]$ for the $n$-element set $\left\{1,2,\ldots,n\right\}$, and the notation $S_n$ for the group of permutations of this set.)
Solution. Let $\sigma\in S_{n}$. Thus, $\sigma$ is a bijection from $\left[ n\right] $ to $\left[ n\right] $. Hence, we can substitute $i$ for $\sigma\left( i\right) $ in the sum $\sum_{i\in\left[ n\right] } \sigma\left( i\right) $. We thus obtain \begin{equation} \sum_{i\in\left[ n\right] }\sigma\left( i\right) =\sum_{i\in\left[ n\right] }i. \label{sol.perm.roundtable.a.sum} \tag{1} \end{equation}
For each integer $z$, we let $z\%n$ denote the remainder of $z$ when divided by $n$.
Assume (for the sake of contradiction) that the $n$ numbers \begin{equation} \left( \sigma\left( 1\right) -1\right) \%n,\qquad\left( \sigma\left( 2\right) -2\right) \%n,\qquad\ldots,\qquad\left( \sigma\left( n\right) -n\right) \%n \label{sol.perm.roundtable.a.nums} \tag{2} \end{equation} are distinct. Thus, the map \begin{align} \left[ n\right] \rightarrow\left\{ 0,1,\ldots,n-1\right\} ,\qquad i\mapsto\left( \sigma\left( i\right) -i\right) \%n \end{align} (this map is well-defined, because for each $i\in\left[ n\right] $, the remainder $\left( \sigma\left( i\right) -i\right) \%n$ belongs to $\left\{ 0,1,\ldots,n-1\right\} $) is injective. Therefore, by the Pigeonhole Principle for Injections, this map must also be bijective (since it is an injective map between two finite sets of the same size). In other words, it is a bijection. Hence, we can substitute $\left( \sigma\left( i\right) -i\right) \%n$ for $j$ in the sum $\sum_{j\in\left\{ 0,1,\ldots,n-1\right\} }j$. We thus find \begin{align} \sum_{j\in\left\{ 0,1,\ldots,n-1\right\} }j=\sum_{i\in\left[ n\right] }\underbrace{\left( \left( \sigma\left( i\right) -i\right) \%n\right) }_{\substack{\equiv\sigma\left( i\right) -i\operatorname{mod} n\\\text{(because }z\%n\equiv z\operatorname{mod}n\\\text{for each integer }z\text{)}}}\equiv\sum_{i\in\left[ n\right] }\left( \sigma\left( i\right) -i\right) =\sum_{i\in\left[ n\right] }\sigma\left( i\right) -\sum _{i\in\left[ n\right] }i=0\operatorname{mod}n \end{align} (by \eqref{sol.perm.roundtable.a.sum}). In view of \begin{align*} \sum_{j\in\left\{ 0,1,\ldots,n-1\right\} }j & =0+1+\cdots+\left( n-1\right) =\dfrac{\left( n-1\right) \left( \left( n-1\right) +1\right) }{2}\qquad\left( \text{by Little Gauss}\right) \\ & =\dfrac{\left( n-1\right) n}{2}, \end{align*} this rewrites as $\dfrac{\left( n-1\right) n}{2}\equiv0\operatorname{mod}n$. In other words, $n\mid\dfrac{\left( n-1\right) n}{2}$. In other words, $\dfrac{\left( n-1\right) n}{2}/n$ is an integer.
But $\dfrac{\left( n-1\right) n}{2}/n=\dfrac{n-1}{2}$ is not an integer, because $n-1$ is odd (since $n$ is even). This contradicts the fact that $\dfrac{\left( n-1\right) n}{2}/n$ is an integer.
This contradiction proves that our assumption was false. Hence, the $n$ numbers listed in \eqref{sol.perm.roundtable.a.nums} are not all distinct. In other words, there exist two distinct elements $i$ and $j$ of $\left[ n\right] $ such that $\left( \sigma\left( i\right) -i\right) \%n=\left( \sigma\left( j\right) -j\right) \%n$. These $i$ and $j$ must then satisfy \begin{align*} \sigma\left( i\right) -i & \equiv\left( \sigma\left( i\right) -i\right) \%n\qquad\left( \text{since }z\equiv z\%n\operatorname{mod}n\text{ for each integer }z\right) \\ & =\left( \sigma\left( j\right) -j\right) \%n\\ & \equiv\sigma\left( j\right) -j\operatorname{mod}n\qquad\left( \text{since }z\%n\equiv z\operatorname{mod}n\text{ for each integer }z\right) . \end{align*} Thus, we have shown that there exist two distinct elements $i$ and $j$ of $\left[ n\right] $ such that $\sigma\left( i\right) -i\equiv\sigma\left( j\right) -j\mod n$. This solves the exercise. $\blacksquare$
(See also Permuting elements of a set around a circle for a similar but more advanced variant of this problem.)