05AB1E, 11 bytes

;3ss{Ƶ}]\\^

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The core program without the restricted-source would have been 2 bytes:

;^

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Explanation:

Explanation of the core program:

;    # Halve the (implicit) input
 ^   # Implicitly truncate it to an integer, and bitwise-XOR it with the
     # (implicit) input
     # (after which the result is output implicitly)

The idea is to add no-ops between them without changing the functionality. I started by creating a generator to get all 05AB1E character-pairs that differ by a single bit. Before the ^, only \ and V seemed reasonable candidates (\ discards the top of the stack; and V pops the top of the stack and saves it in variable Y). From there it was kinda trial-and-error to see what characters I could use. Because the ; and ^ differ by 4 bits, at least that many no-op bytes in between are necessary. Although I have the feeling the byte-count can still be improved a bit perhaps, 9 bytes to change 4 specific bits without changing the functionality of the core program isn't too bad I guess.

Explanation of the new program:

;           # Halve the (implicit) input-integer
            # No-ops:
 3          #  Push 3
  s         #  Swap the two values on the stack
   s        #  Swap them back
    {       #  Sort the digit(s) of 3
     Ƶ}     #  Push compressed integer 226
       ]    #  Close all open if-statements and loops
        \   #  Discard the top of the stack (the 226)
         \  #  Discard the top again (the 3)
          ^ # Implicitly truncate the halved input to an integer, and bitwise-XOR
            # it with the (implicit) input
            # (after which the result is output implicitly)

(See this 05AB1E tip (section How to compress large integers?) to understand why Ƶ} is 226.)

The bytes in 05AB1E's codepage of ;3ss{Ƶ}]\\^ are:

;  00111011
       v
3  00110011
    v
s  01110011
s  01110011
       v
{  01111011
        v
Ƶ  01111111
         v
}  01111101
     v
]  01011101
          v
\  01011100
\  01011100
         v
^  01011110