f(x) = x1*x2 + x3*x4 mod 2 (x1,x2,x3,x4 are the four input bits, x is their concatenation)
e is the changing bits, x is the possible inputs, the last column p calculates the the number of different inputs where the output changes if e is added to x before inputting it to f divided by the total number of inputs possible. This is called a probability. Each cell is the output of f(x + e). The first row where e=0000 is just the plain output of f for every input x. Changes are calculated in respect to this row. According to the general SAC (Avalanche Criteria, for 1 Bit and also for higher orders up to 4 bits) this should be 50 for.

 e\x | 0000 | 0001 | 0010 | 0011 | 0100 | 0101 | 0110 | 0111 | 1000 | 1001 | 1010 | 1011 | 1100 | 1101 | 1110 | 1111 | p
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0000 |  0   |  0   |  0   |  1   |  0   |  0   |  0   |  1   |  0   |  0   |  0   |  1   |  1   |  1   |  1   |  0   |  -
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0001 |  0   |  0   |  1   |  0   |  0   |  0   |  1   |  0   |  0   |  0   |  1   |  0   |  1   |  1   |  0   |  1   | 8/16
0010 |  0   |  1   |  0   |  0   |  0   |  1   |  0   |  0   |  0   |  1   |  0   |  0   |  1   |  0   |  1   |  1   | 8/16
0011 |  1   |  0   |  0   |  0   |  1   |  0   |  0   |  0   |  1   |  0   |  0   |  0   |  0   |  1   |  1   |  1   | 8/16
0100 |  0   |  0   |  0   |  1   |  0   |  0   |  0   |  1   |  1   |  1   |  1   |  0   |  0   |  0   |  0   |  1   | 8/16
0101 |  0   |  0   |  1   |  0   |  0   |  0   |  1   |  0   |  1   |  1   |  0   |  1   |  0   |  0   |  1   |  0   | 8/16
0110 |  0   |  1   |  0   |  0   |  0   |  1   |  0   |  0   |  1   |  0   |  1   |  1   |  0   |  1   |  0   |  0   | 8/16
0111 |  1   |  0   |  0   |  0   |  1   |  0   |  0   |  0   |  0   |  1   |  1   |  1   |  1   |  0   |  0   |  0   | 8/16
1000 |  0   |  0   |  1   |  0   |  1   |  1   |  0   |  0   |  0   |  0   |  0   |  1   |  0   |  0   |  0   |  1   | 8/16
1001 |  0   |  0   |  1   |  0   |  1   |  1   |  0   |  1   |  0   |  0   |  1   |  0   |  0   |  0   |  1   |  0   | 8/16
1010 |  0   |  1   |  0   |  0   |  1   |  0   |  1   |  1   |  0   |  0   |  1   |  0   |  0   |  1   |  0   |  0   | 8/16
1011 |  1   |  0   |  0   |  0   |  0   |  1   |  1   |  1   |  1   |  0   |  0   |  0   |  1   |  0   |  0   |  0   | 8/16
1100 |  1   |  1   |  1   |  0   |  0   |  0   |  0   |  1   |  0   |  0   |  0   |  1   |  0   |  0   |  0   |  1   | 8/16
1101 |  1   |  1   |  0   |  1   |  0   |  0   |  1   |  0   |  0   |  0   |  1   |  0   |  0   |  0   |  0   |  0   | 8/16
1110 |  1   |  0   |  1   |  1   |  0   |  1   |  0   |  0   |  0   |  1   |  0   |  0   |  0   |  1   |  0   |  0   | 8/16
1111 |  0   |  1   |  1   |  1   |  1   |  0   |  0   |  0   |  1   |  0   |  0   |  0   |  1   |  0   |  0   |  0   | 8/16

As you can see, this function f satisfies the higher order SAC up to the maximum of 4 Bits. You can use four of these functions in parallel or in some other combined way to achieve this also for a 4 Bit to 4 Bit mapping instead of 4 Bit to 1.