Interleaving bytes to make an effectively larger block size Will applying a 16 byte bit block cipher such as AES over a 256 byte block to in the following way result in a much stronger cipher:-


Apply block cipher with first key digest to each 16 byte group.
Then interleave the bytes.  I.E. Place the bytes horizontally into a 2D (16 times 16) byte array and then remove the bytes vertically so that each new 16 byte group now has one byte of each of the old 16 byte groups.
Then apply block cipher with second key digest. 
Then interleave.
Then apply block cipher with third key digest.
Alternative 1, Interleave = ShiftRows


Use a KDF for extending and splitting the input key into three keys $K_1, K_2, K_3$.
Split the input into $A_0 = a_0..a_{15}, A_1 = a_{16}..a_{31},...,A_{16} = a_{240}..a_{255}$
AES-ECB encrypt using key $K_0$ so that $E_{K_0}(A_i) = B_i = b_{16i+0}..b_{16i+15}$
Interleave so that $c_{16i+j} = b_{16(i+j)+j \bmod 256}$


The problem here is that if $A_i = A_j$ for all $0 \le i, j \lt 16$, then $C_i = C_j$ for all $0 \le i, j \lt 16$. This equality will propagate through all three rounds, making it easy to distinguish your 2048-bit composite function from a random 2048-bit permutation with only a single query.

Clearly, if $A_i = A_j$, then $E_{K_0}(A_i) = E_{K_0}(A_j) = B_i = B_j$. This means that $b_{16i+k} = b_{16j+k}$ for all $0 \le k \lt 16$. If this relation holds for all $0 \le i, j \lt 16$ then, for each $0 \le k \lt 16$, $b_{16i + k} = b_{16j + k}$ for all $0 \le i,j \lt 16$. 

In particular, $b_{16i+j} = b_{16(i+j)+j \bmod 256}$. 

In other words, if $A_i = A_j$ for all $0 \le i, j \lt 16$, then the interleave step will be the identity function. The composite function will for such inputs only consist of three consecutive AES-ECB operations.

Alternative 2, Interleave = Transpose


Use a KDF for extending and splitting the input key into three keys $K_1, K_2, K_3$.
Split the input into $A_0 = a_0..a_{15}, A_1 = a_{16}..a_{31},...,A_{16} = a_{240}..a_{255}$
Round = 0
AES-ECB encrypt using key $K_0$ so that $E_{K_0}(A_i) = B_i = b_{16i+0}..b_{16i+15}$
If round = 2, exit
Interleave so that $c_{16i+j} = b_{16j+i}$
round = round + 1, $A = C$ jump to step 4.


In this case, if $A_i = A_j$ for all $0 \le i,j \lt 16$, then, in the first round = 0, $c_{16i+j} = c_{16i + k}$ for all $0 \le i,j,k \lt 16$. This means that after the second round = 1, there are only $2^8$ different values for the $B_i$ blocks.

Hence, find two 128 bit values $A$ and $A'$ such that $E_{K_0}(A) = b_0..b_{15}$ and $E_{K_0}(A') = b'_0..b'_{15}$ differ only in a single byte $b_i \neq b'_i$, and such that $E_{K_1}(b_i..b_i) = c_0..c_{15}$ and $E_{K_1}(b'_i..b'_i) = c'_0..c'_{15}$ are equal in two different byte positions $c_j = c'_j$ and $c_k = c'_k$. If this happens, the two inputs to the composite function that consist in sequences of $A$ and $A'$ respectively, will result in two output, such that both the $j$th blocks and the $k$th blocks are identical. The probability two random 128 bit blocks will have this relation is $\frac{16}{2^{8\times15}}\times\frac{16\times15}{2^{8\times2}}$, which is significantly greater than the probability that you would get the same relation between the corresponding outputs of a 2048 bit pseudo random permutation, and great enough to expect to get such a relation after only $2^{64}$ attempts.