Can you quarter the processing time for real, symmetric FFTs? The DFT of a real signal is Hermite-symmetric, so you can roughly halve the computation time/memory by not bothering to calculate half the values of the spectrum (and complex conjugating the existing values and copying them to the second half if needed).  So the rfft operation takes N samples and outputs N/2 spectrum bins in half the time, for instance.

Signals which are even-symmetrical and real have even-symmetrical and real spectra (and real spectra take half the memory to store as complex spectra), so for symmetrical input, can the calculation be halved again by only using the first half of the signal (N/2) to generate the first half of the spectrum (N/2)?  How?

Is there a way to take the regular FFT of the N/2 half-signal and manipulate the output to produce the N/2 half-spectrum?

(real/odd ⇔ imaginary/odd would work, too, but real/even ⇔ real/even case is simpler to follow.)
I'm not sure about quartering the process time but this is what I can think of for now. N point DFT of x.

$X_k = \sum_{n=0}^{N-1}x_nW_N^{kn}$

By the even symmetry assumption

$X_k = \sum_{n=0}^{N/2-1}x_n(W_N^{kn}+W_N^{k(N-1-n)})$

Then we simplify

$X_k = W_N^{k(N-1)/2}\sum_{n=0}^{N/2-1}x_n(W_N^{kn}W_N^{-k(N-1)/2}+W_N^{-kn}W_N^{k(N-1)/2})$

$X_k = 2W_N^{k(N-1)/2}\sum_{n=0}^{N/2-1}x_n\cos(2\pi k\frac{n-(N-1)/2}{N})$

There are probably some errors but if simplifications are valid the 'FFT' size is reduced by a factor of 2 and the twiddle factors are real (this maybe corresponds to your quartering). As in normal FFT the symmetry and periodicity can be exploited to decompose the 'symmetric FFT' into radix-2 steps for instnace.