The questions below are motivated by pure curiosity. I heard of the first question from my former advisor. I have no idea how difficult they are, since I have no experience with magic squares.

By a normal magic square of order $n$ I mean a $n\times n$ magic square whose terms are all of the numbers $0,1,\ldots,n^2-1$.

Is it possible to construct an infinite sequence $M_n$ of normal magic squares such that $M_{n}$ is a block submatrix of $M_{n+1}$ lying in the centre of $M_{n+1}$ (i.e. to obtain $M_{n}$ we remove from $M_{n+1}$ the $k$ top rows, $k$ bottom rows, $k$ columns from the left and $k$ columns from the right)?

Can one construct a normal magic square of odd order (greater than $1$) with $0$ as the central element of the square?

Note that a positive answer to the second question gives a positive answer to the first one. Let $A=[a_{ij}]_{0\leq i,j\leq n}$ be a magic square as in question 2. For a number $c$ we shall use the notation $A+c=[a_{ij}+c]_{0\leq i,j\leq 2n}$ Then: $$A'=\left[\begin{matrix}A+(n+1)^2a_{00} & \dots & A+(n+1)^2a_{0n} \\ \vdots & \ddots & \vdots\\ A+(n+1)^2a_{n0} & \dots & A+(n+1)^2a_{nn}\end{matrix}\right]$$ is a normal magic square with $A$ in the centre.