full version is here https://cybersecctf.github.io/blog/?q=shufflebox%20downunderctf2024

we see provided py coded it create random 16 length permutation that change s1 and s2 and s3 to t1 and t2 and t3 on base of txt provided file but
s3 is not known and should find it with this code so use this code

#python
import itertools


def find_all(string, char):
    return [i for i, x in enumerate(string) if x == char]

def solve(s1, t1, s2, t2, t3=None):
    # Collect positions for each character
    positions = {}
    for char in 'abcd':
        positions[char] = find_all(t1, char)

    # Generate all possible permutations of positions
    pos = [p1 + p2 + p3 + p4 for p1 in itertools.permutations(positions['a'])
                                for p2 in itertools.permutations(positions['b'])
                                for p3 in itertools.permutations(positions['c'])
                                for p4 in itertools.permutations(positions['d'])]

    # Check if permutation that works for s1 works for s2
    valid_perms = []
    for perm in pos:
        if ''.join(t2[perm[i]] for i in range(16)) == s2:
            valid_perms.append(perm)

    assert len(valid_perms) == 1

    # Reverse permutation for t3 or s3
    if t3 is not None:
        result = ''.join(t3[valid_perms[0][i]] for i in range(16))
    elif s3 is not None:
        result = ''.join(s3[valid_perms[0][i]] for i in range(16))
    else:
        raise ValueError("Either t3 or s3 must be provided.")

    return result

# Example usage: 
s1 = "aaaabbbbccccdddd"
t1 = "ccaccdabdbdbbada"
s2 = "abcdabcdabcdabcd"
t2 = "bcaadbdcdbcdacab"
t3 = "owuwspdgrtejiiud"
print( solve(s1, t1, s2, t2, t3))

and get flag after run it and wrap it with ductf

Original writeup (https://cybersecctf.github.io/blog/?q=shufflebox%20downunderctf2024).