Solution:

We will solve a slightly more general problem: suppose the chessboard can have any number $N$ of squares, not necessarily a square number. We will show that the magicians have a strategy if and only if $N$ is a power of two. In the particular case $N = n^{2}$, it follows that $n$ must be a power of two.

If there is a strategy, then $N$ is a power of two. Let us count the number $D_{c}$ of checker arrangements that the second magician associates with a given square $c$. From every possible initial arrangement the first magician can reach one of these by means of exactly one change, so every possible arrangement must be reachable starting from one of the arrangements associated with $c$ and changing exactly one checker. Since there are $N$ possible checker changes, $D_{c} \geq 2^{N} / N$. On the other hand the sum of the numbers $D_{c}$ as $c$ ranges over the $N$ possible squares is less than or equal to $2^{N}$. It follows that $D_{c} = 2^{N} / N$ for every square $c$, so $N$ divides $2^{N}$.

If $N$ is a power of two, then there is a strategy. For $N = 1$ the second magician cannot be wrong. For $N = 2$, let us call the two squares $x$ and $y$: the magicians agree to have on square $x$ a white checker if Daniele chose $x$, and a black checker if Daniele chose $y$. Now, it suffices to show that if the magicians have a strategy for $N = a$ and for $N = b$, then they have a strategy for $N = ab$. The strategy consists in imagining the squares arranged in an $a \times b$ grid - with $a$ rows and $b$ columns - and stipulating that a row or column is white if it contains an even number of white checkers, black otherwise. In this way, by changing the color of a checker, exactly the color of the row and of the column on which it lies changes. The first magician determines the row and the column on which the square chosen by Daniele lies, and uses the strategies for $N = a$ and $N = b$ on the colors of the rows and of the columns respectively, then decides to change the color of exactly one row and one column, and replaces the checker that is at their intersection. The second magician considers the colors of the rows and of the columns, and by applying the strategies of the cases $N = a$ and $N = b$ determines on which row and on which column the square chosen by Daniele must lie.