Shouldn't Charge Conjugation be known as "positive/negative frequency symmetry"? I know that charge conjugation exchanges the creation (or annihilation) operators of the particles with those of the anti-particles and therefore merits the name charge conjugation. 
However, if operated on the single electron Dirac plane wave $u(p)$ it results in v(p) and vice-versa. For me, however, $v(p)$ is not the single positron plane wave. For me it is the negative frequency solution. So for the single particles solutions of the Dirac equation it is more like a symmetry between positive and negative solutions.
For a charge conjugation operator I would expect that it changes a in-going single electron plane wave to a in-going single positron wave. But $v(p)$ represents a out-going plane wave in Feynman diagrams.
It is also said that $C$ changes the negative frequency wave $v(p)$ to a positive frequency wave solution $u(p)$ which finally represents the positron.
Okay, but again then $C$ should not be called a charge conjugation,
but symmetry between positive and negative frequency solutions.
I would be grateful to get an explanation on that. 
 A: When investigating spinorial representations of the Lorentz group, one finds that if $\Psi$ is a left-handed Dirac spinor, then $\Psi^c = -i\gamma^2\Psi^*$ is a right-handed Dirac spinor. At that moment, however, the physical meaning of the operation is latent.
Having quantized the Dirac spinor,
$$
\Psi(x) = \int \frac{d^3p}{(2\pi)^3\sqrt{2E_p}} \sum_{s=1,2} \left(a_{p,s}u^s(p)e^{-ipx} +  b^\dagger_{p,s}v^s(p)e^{+ipx} \right)\\
\propto
\left(
a_{p,1}u^1(p)e^{-ipx}  
+a_{p,2}u^2(p)e^{-ipx}
+b^\dagger_{p,1}v^1(p)e^{+ipx}  
+b^\dagger_{p,2}v^2(p)e^{+ipx}
\right)
$$
we reconsider the meaning of $\Psi^c$. By brute force we find that the one-particle spinors obey,
$$
[-i\gamma^2u^1(p)]^* = v^2(p),\\
[-i\gamma^2u^2(p)]^* = -v^1(p),\\
[-i\gamma^2v^1(p)]^* = -u^2(p),\\
[-i\gamma^2v^2(p)]^* = u^1(p).
$$
Because of these transformations properties, we guess that the creation and annihilation operators transform in an analogous fashion, e.g.
$$
Ca_{p,1}C = \eta_c b_{p,2},
$$
note well, however, that
$$
Cu^s(p)C = u^s(p),
$$
etc. With all these formula, you can show that,
$$
C\Psi(x)C \propto \eta_c
\left(
b_{p,2}u^1(p)e^{-ipx}  
-b_{p,1}u^2(p)e^{-ipx}
-a^\dagger_{p,2}v^1(p)e^{+ipx}  
+a^\dagger_{p,1}v^2(p)e^{+ipx}
\right)\\
=
-i\eta_c\gamma^2\Psi^*
$$
Justifying our guesses. 
Lastly, we check the consequences of the transformation properties of the creation and annihlation operators. Looking at e.g. a $U(1)$ charge,
$$
Q \propto a^\dagger a - b^\dagger b
$$
it's clear that $CQC=-Q$, justifying the name charge-conjugation.
