Interleaving bytes to make an effectively larger block size
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.