Are the FFT coefficients symmetric in image processing?
For real images, there is indeed a formal redundancy, termed Hermitian or conjugate-symmetric as detailed by @Fat32. This symmetry however is "modulated" by the complex expression of the Fourier coefficients. So the FFT requires half the number of coefficients, but twice the amounts, due to the real/imaginary or modulus/phase couples. All in all, the redundancy vanishes. 

More precisely, an FFT on a $2n$ length signal provides you with $n+1$ non redundant coefficients: the first (DC) and the last (Nyquist) are always real, the $n-1$ often complex. So you have $n-1$ complex coefficients, plus one pair of reals, altogether $n$ "pairs" of coefficients. 

In general, with linear transformations, you cannot expect a bijective, one-to-one mapping with less transformed coefficients than samples in the signal/image, unless, as detailed by @Fat32, you reduce the data space, but then again, the transformed space would naturally have the same dimension as that of the data.