On the use of Permutation matrix to perform iFFT I have a question about using the permutation matrix for performing $iFFT$ for such matrix and then reshape it row-wise and column-wise way. 

Let's say that we have a random matrix $x$ whose size is (32,8), so that matrix can be written as below: 
 

where $x_k$ represents the column $k$ of matrix $x$ with size of 32 for each column. 

$1$- Let's take column wise $iFFT$ for matrix $x$ and then reshape it row-wise resulting matrix $Y$ as below matlab code 

x = randn(32,8); 
y = ifft(x); 
Y = reshape(y.',[],1);


$2$- let's manipulate that using another way of permutation matrix. first we can reshape the matrix $x$ column-wise into (256,1) as below: 



Then performing $iFFT$ operation with size of $32$-point is corresponding to: 


$F$ is the Fourier transformation matrix
In order to get the equivalent matrix of Y representing the reshape of $iFFT(x)$ in row-wise, we can multiply the matrix Z with permutation matrix P resulting vector $Y2$ which should be similar to vector $Y$ in case $1$. That can be done by matlab as: 

z = reshape(x,[],1);   %
X = kron(eye(8),dftmtx(32)); 
Z = X*z;   

P = zeros(m*n);       %Building the permutation matrix P
col = 1;
for i = 1:m
    for j = 1:n
        E = zeros(m,n);
        E(i,j) = 1;
        P(:,col) = E(:);
        col = col + 1;
    end
end

Y2 = P*Z


So, for $(1)$ and $(2)$,   $Y$ should be similar to $Y2$. Is that right? but what I get is different results!  My question is that calculation right? why do I get different results ?
Y and Z should show the same values

clear;
%Create random vector x 
x = randn(32,8); 

%take ifft along each column 
y = ifft(x); 
Y = reshape(y,[],1);

z = reshape(x,[],1);   %
d=conj(dftmtx(32))/32;
X = kron(eye(8),d); 
Z = X*z;