Is it possible to reduce multiple permutation tables to a single table?

Question

Apologies if this is not the best place to post this, but there are permutations and reductions involved and it's not the programming that I'm interested in. If not I'd be happy for this question to be transferred somewhere else more appropriate. I don't really know how to phrase this other than by pseudo computer code.

Let's consider the Pearson hash function, which I describe in pseudo code:-

algorithm pearson hashing is
    h := 0

    for each c in C loop
        h := T[ h xor c ]
    end loop

    return h

where T is a permutation such as:-

    T = [1, 87, 49, 12, 176, 178, 102, 166, 121, 193, 6, 84, 249, 230, 44, 163,
         14, 197, 213, 181, 161, 85, 218, 80, 64, 239, 24, 226, 236, 142, 38, 200,
         110, 177, 104, 103, 141, 253, 255, 50, 77, 101, 81, 18, 45, 96, 31, 222,
         25, 107, 190, 70, 86, 237, 240, 34, 72, 242, 20, 214, 244, 227, 149, 235,
         97, 234, 57, 22, 60, 250, 82, 175, 208, 5, 127, 199, 111, 62, 135, 248,
         174, 169, 211, 58, 66, 154, 106, 195, 245, 171, 17, 187, 182, 179, 0, 243,
         132, 56, 148, 75, 128, 133, 158, 100, 130, 126, 91, 13, 153, 246, 216, 219,
         119, 68, 223, 78, 83, 88, 201, 99, 122, 11, 92, 32, 136, 114, 52, 10,
         138, 30, 48, 183, 156, 35, 61, 26, 143, 74, 251, 94, 129, 162, 63, 152,
         170, 7, 115, 167, 241, 206, 3, 150, 55, 59, 151, 220, 90, 53, 23, 131,
         125, 173, 15, 238, 79, 95, 89, 16, 105, 137, 225, 224, 217, 160, 37, 123,
         118, 73, 2, 157, 46, 116, 9, 145, 134, 228, 207, 212, 202, 215, 69, 229,
         27, 188, 67, 124, 168, 252, 42, 4, 29, 108, 21, 247, 19, 205, 39, 203,
         233, 40, 186, 147, 198, 192, 155, 33, 164, 191, 98, 204, 165, 180, 117, 76,
         140, 36, 210, 172, 41, 54, 159, 8, 185, 232, 113, 196, 231, 47, 146, 120,
         51, 65, 28, 144, 254, 221, 93, 189, 194, 139, 112, 43, 71, 109, 184, 209]

and xor is the Exclusive OR function. So for example a Pearson hash of "Hello" might return 42. What I want to focus on is table T, which is a random permutation on the range (0, 255). Let me now add more independent permutations (T2, T3 and T4) into the loop:-

for each c in C loop
    h := T1[ h xor c ]
    h := T2[ h xor c ]
    h := T3[ h xor c ]
    h := T4[ h xor c ]
end loop

So we have four different permutation tables, T1, T2, T3 and T4 and the value h passes through all four sequentially.

Q. Is it possible to reduce all four tables to one single new one (Tnew) that produces the same output, as:-

for each c in C loop
    h := Tnew[ h xor c ]
end loop

?