Sign extension. Your compiler is implementing char as a signed char. When you pass the chars to printf they are all being sign extended during their promotion to ints. When the first bit is a 0 this doesn’t matter, because it gets extended with 0s.
0xAF in binary is 10101111 Since the first bit is a 1, when passing it to printf it is extended with all 1s in the conversion to int making it 11111111111111111111111110101111, which is 0xFFFFFFAF, the hex value you have.
Solution: Use unsigned char (instead of char) to prevent the sign extension from occurring in the call
const unsigned char raw[] = {0x20,0x00,0xAF,0x00,0x69,0x00,0x33,0x00,0x5A,0x00};
All of these values in your original example are being sign extended, it’s just that 0xAF is the only one with a 1 in the first bit.
Another simpler example of the same behavior (live link):
signed char c = 0xAF; // probably gives an overflow warning
int i = c; // extra 24 bits are all 1
assert( i == 0xFFFFFFAF );