Am I violating Nyquist?
Yes you are violating! But not the Nyquist rate as a matter of fact :))

First of all Nyquist (Shannon) sampling theorem is based on the analog signal's bandwidth, and states that you must sample the signal at a rate twice the bandwidth of it. Now, what's the bandwidth of your signal? You must compute this one first.

Given a DAC (digital to analog converter) modeled as D/C block; i.e., an ideal signal reconstructor with an ideal interpolation (image rejection) filter with $h_r(t)$ whose cutoff frequency is at a half of the DAC sampling rate, then the output of the DAC given at a sampling rate of $1$G samples per second (1 GHz) will be bandlimited to $500$ MHz.

Now given a continuous-time (analog) bandlimited baseband signal $x(t)$ with a bandwidth of $500$ MHz, the required Nyquist sampling rate will be twice the bandwidth of the signal which is $1$ GHz, or stated as $1$ G samples per second.

Hence in principle (ideally) there's nothing wrong with (Nyquist is not violated by) using a DAC-ADC communication pair with the same sampling frequency. Of course due to several nonidealities (please refer to @MarcusMüller 's answer for those) you will find it quite impractical to achieve. Anyway, Nyquist is ok ;-)

But, then, you may be confusing the sampling theorem with channel capacity (information rate) for AWGN channel. So given your DAC-ADC AWGN channel system with a bandwidth of $B = F_s/2$, and given an SNR of the line in between; your maximum rate of information transfer is given by Shannon-hartley law which states that $C = B \times \log_2( 1 + SNR)$ b/s...

Assuming an SNR of $90$ dB (for about $16$ bits of an ADC,DAC with LSB noise) this means in principle you should be able to reach a data rate of $C = 500 \times 10^6 \times 15$ bits per second. This is about $7.5$ Gbps.