Problems with IFFT not being symmetrical
Here is an example using Matlab/Octave code. There is no need to conjugate.

x = rand(1000,1);
y = rand(1000,1);
z = conv(x,y,'full');
y2 = ifft( fft(z,1999) ./ fft(x,1999) );
norm(y - y2(1:1000)) / norm(y)

ans =

   1.5654e-13