Steven is right, but I would add some detail about the theory. I've also written something about this in an answer about FFT.
The short answer is that when computing the spectrum of a signal, we are applying the Fourier transform (or something fancier, but let's stay with that). What happens is that every signal that is discrete in time, will be periodic in frequency, and vice versa.
For the same reason, if you have a perfect sinewave, it will be a single spike in the spectrum.
A consequence of this phenomenon is aliasing: these copies of the spectrum are separated by the sampling frequency (the frequency at which the signal is measured); if it's too low (less than 2*fmax) these copies will merge and you'll lose detail about the original signal. This is the Shannon's theorem.
You're on the right track but you need to get your transfer function into standard form.
Gathering terms and multiplying through, we get:
\$\dfrac{V_o}{V_i} = H(j\omega) = \dfrac{1 - j\omega R_6 C_1}{1 + j\omega R_6 C_1}\$
This is an all-pass filter with a magnitude response of 1. To verify, multiply by the complex conjugate to get \$|H(j \omega)|^2\$:
\$|H(j\omega)|^2 = H H^* = \dfrac{1 - j\omega R_6 C_1}{1 + j\omega R_6 C_1} \dfrac{1 + j\omega R_6 C_1}{1 - j\omega R_6 C_1} = 1\$
For the phase, multiply top and bottom by \$(1 - j\omega R_6 C_1)\$:
\$H(j \omega) = \dfrac{[1 - (\omega R_6 C_1)^2] - j2\omega R_6 C_1}{(1 + \omega R_6 C_1)^2}\$
In this form, the real and imaginary parts are manifest and, for \$\angle H(j \omega) = -90^\circ\$, the real part must be zero.
\$\angle H(j \omega) = \tan^{-1}\frac{-2\omega R_6 C_1}{1 - (\omega R_6 C_1)^2} \$
However, for this particular transfer function (in the first form I wrote), it's easy to see that the numerator and denominator have equal and opposite phase. Thus, the phase of this transfer function is simply twice the phase of the numerator. Then, by inspection, we see that \$\angle H(j \omega) = -90^\circ\$ when \$\omega = \frac{1}{R_6 C_1}\$.
(Thanks to Szymon Bęczkowski for the nice plots!)
Best Answer
That's what they call a logarithmic scale
https://en.wikipedia.org/wiki/Logarithmic_scale
multiples have the same distance on a logarithmic scale eg: doubling: the distance 30 to 60 matches the distance 40 to 80 and 100 to 200 etc.
logarithmic scales are useful where arithmetic ratios are more significant than arithmetic differences.