Comparability of elements in a Latin square based on a few rows

Question

Let $\Pi=\{\pi_1,\pi_2,\dots,\pi_n\}$ be the rows of an $n\times n$ Latin square (the order of the rows does not matter).

Each row $\pi_i$ induces an order $\prec_i$ on the elements of $[1,n]$, where $a \prec_i b$ iff $a$ appears on the left of $b$ in $\pi_i$.

The question: For all $\Pi$, is there a small subset of rows $\Pi' \subseteq \Pi$ such that for any two element $a,b \in [1,n]$, we have both $a \prec_i b$ and $b \prec_j a$ for some rows $\pi_i$ and $\pi_j$ in $\Pi'$? In particular, is the minimum size of $\Pi'$ upper bounded by a constant or by a function growing slower than $\log n$?

(Optional:) If the answer is negative, I would like to relax the order to $\preceq_i$, where $a \preceq_i b$ iff $a$ appears on the left of $b$ or $a$ is adjacent to $b$ in $\pi_i$ (i.e. possibly just after, but not farther). Would the answer change in this case?

Answer 1

The answer to the first question is that for some Latin squares of order $n$ (namely the addition table of a cyclic group) the minimum size of $\Pi'$ is $n$.

With addition modulo $n$, let $\pi_i=(i,i+1,i+2,\dots,i+n-1)$ for $i\in[0,n-1]$. If $a\in[0,n-1]$ then $a\prec_i a-1$ only when $i=a$. Hence we must have $\Pi'=\Pi$.

For the second question, the same example shows that $\Pi'$ must have at least $\lceil n/2\rceil$ elements, since $a\preceq_i a-2$ only when $i\in\{a,a-1\}$.