There seems to be one important property to a magic square (ms):
the sum of all squares devided by the number of squares is equal to the center-number
As such, your center-4 magic square would need a sum of 36 and your center-6 magic square would need a sum of 54. Added together this would be a sum of 90 - so far it seems possible. The 4-ms would need a row/column/diagonal sum of 12 and the 6-ms would need one of 18. It follows, that the 4-ms needs all the 1s and 2s, the 6-ms on the other hand needs all the 9s and 8s. Looking at the 4ms: 4 + 1 + x = 12 -> the 4ms needs a 7 for every 1 it has. Same goes for 4+2+x=12 -> it needs a 6 for every 2 it has. And this is the point where it stops working, since you only have 2 6s.
So as long as the mentioned property is mandatory, it doesn´t work. I wasn´t able to proof/find proof yet, that it is mandatory. It just seems intuitive by looking at proper magic squares. But should you be able to find proof that it is mandatory, it would equal proof that your initial idea doesn´t work.