Q:

Construct $4 \times 4$ magic square with fixed "1"

The method I have found to generate $4\times 4$ magic squares gives me a result in which the number "1" is at of the corners of the square.  How can we extend this to a method to generate a magic square, for a fixed location of number "$1$"? 
The number "$1$" can be in $16$ different locations (cells).  If we name the cells, from upper left corner: $1, 2, 3,4, 5, \ldots,$ and the number "$1$" is at the $i^{th}$ cell, then how we can fill the other cells to make a magic square? 
You can see variations here. 

A:

There are really only three different locations:  a corner, a side next to a corner, and one step diagonally in from a corner.   If we can put $1$ into each of these, we can put it in any cell by using rotations and reflections.   We already have a corner.   We can subtract every number from $16$ and keep the square magic, but that doesn't help because it puts $1$ in the lower right corner.  Because of the arrangement, we can add $8$ to all the numbers below $8$ and subtract $8$ from all those above, giving
$$\begin {array} {c|c|c|c} 9&7&6&12 \\ \hline 4&14&15&1 \\ \hline  16&2&3&13
\\ \hline 5&11&10&8 \end {array}$$
That gets next to the corner.   We can also rotate the $2 \times 2$ blocks by $180^\circ$ to get $$\begin {array} {c|c|c|c} 6&12&9&7 \\ \hline 15&1&4&14 \\ \hline  3&13&16&2
\\ \hline 10&8&5&11 \end {array}$$ which gets one in from the corner

