Impedance matching a capacitor load
Unfortunately, you cannot match a purely reactive load.  You could cancel out the reactance at a particular frequency using an inductor where $$Z_L=j*\omega*L$$ and $$ Z_C=\frac{1}{j*\omega*C}=\frac{-j}{\omega*C}$$

However, that would just give you a short circuit.  In any case using the formula for reflection coefficient of a load on a 50 ohm line $$ \Gamma=\frac{Z_{Load}-50}{Z_{Load}+50}$$

The magnitude of Gamma for any purely imaginary Z_Load will be 1 (complete voltage wave reflection).  For example, if Z_Load is j50, you'll get a Gamma of j.  If Z_load is j25 you'll get a Gamma of -0.6 + j0.8.  If it's -j33 you'll get -0.39-j0.92, etc.

I would just try to keep your transmission line very short and hope your source has sufficiently low output impedance to deal with this very non-ideal load.