Exchanging two columns or rows of a regular square yields another regular square

Question

An $n$-by-$n$ square is regular if the two conditions are met:

1. Each of the integers from $0$ to $n^2 − 1$ appears in exactly one cell, and each cell contains only one integer (so that the square is filled), and
2. If we express the entries in base-$n$ form, each base-$n$ digit occurs exactly once in the units’ position, and exactly once in the $n$’s position.

So if S is a regular square (as defined above), and T is obtained from S by exchanging two rows or two columns, then is T also a regular square? Why is it a regular square?

Answer 1

As you have stated the definition, you can derive item 2 from item 1. Swapping columns or rows doesn't change the set of numbers that are in the cells, so it is still a regular square.

I believe you want item 2 to require that each row and column contain each base-$n$ digit once in the ones place and once in the $n$'s place.

Even so, swapping a pair of rows (or columns) will make $T$ a regular square if $S$ is. If $S$'s rows are proper, so are $T$'s, as they are the same. Each column of $T$ has the same set of entries as the corresponding column of $S$, so it satisfies 2 as well.

Answer 2

I hope it's clear that condition $1$ will still hold.

If you exchange two columns, does condition $2$ still hold for all the columns? (Exchanging columns will keep every column the exact same, but merely put it in a different spot)

If you exchange two columns, does condition $2$ still hold for all the rows? (Note that permuting the columns will induce a permutation on the entries in individual rows, this is key here, convince yourself that it's true!)

To think about what happens when permuting rows, think about the above, but exchange the words "rows" and "columns".