Short answer: it depends on which $4\times 4$ matrix representation is used.
In any representation, as far as its action on a Dirac spinor $\psi$ is concerned, the connected part of the Lorentz group is generated by transformations of the form
$$
\psi\rightarrow
S
\psi
\hskip2cm
S = \exp\left(\frac{\theta}{2}\gamma^\mu\gamma^\nu\right)
\tag{1}
$$
with $\mu\neq \nu$. This is a Lorentz boost if $\mu=0$ or $\nu=0$, and it is an ordinary rotation if $\mu\geq 1$ and $\nu\geq 1$.
Now suppose that we use a representation in which $\gamma^\mu$ is
hermitian for $\mu=0$ and anti-hermitian for $\mu =1,2,3$
symmetric for $\mu=0,2$ and anti-symmetric for $\mu=1,3$
The representation used in Peskin and Schroeder's An Intro to QFT satisfies these conditions. In this representation, the matrix
$$
C\propto \gamma^0\gamma^2
\tag{2}
$$
satisfies
$$
(\gamma^\mu)^T C = -C\gamma^\mu,
\tag{3}
$$
which implies
$$
S^T C = C S^{-1}.
\tag{4}
$$
Therefore, under the Lorentz transformation $\psi\rightarrow S\psi$, the quantity $\psi^T C\psi$ transforms as a scalar:
$$
\psi^T C\psi\rightarrow \psi^T S^T CS\psi = \psi^T C\psi
\tag{5}
$$
and the quantity $\psi^T C\gamma^\mu\psi$ transforms as a vector:
$$
\psi^T C\gamma^\mu\psi\rightarrow \psi^T S^T C\gamma^\mu S\psi
= \psi^T C (S^{-1}\gamma^\mu S)\psi.
\tag{6}
$$
By the way, the similar-looking quantity $\psi^T \gamma^\mu C\psi$
transforms like this:
$$
\psi^T \gamma^\mu C\psi\rightarrow \psi^T S^T \gamma^\mu CS\psi
= \psi^T S^T \gamma^\mu (S^T)^{-1}C\psi,
\tag{7}
$$
which is not the way a vector (or any other tensor) should transform.
In any representation, we can think of (4) as the defining condition for a matrix $C$. The matrix $C$ that satisfies this condition depends on the representation; but if such a matrix exists, then equations (5)-(6) hold automatically. We can think of equations (5)-(6) as the motive for the condition (4).
Similarly, if we choose a matrix $A$ so that
$$
S^\dagger A = A S^{-1},
\tag{8}
$$
then the quantity $\psi^\dagger A\psi$ transforms as a scalar:
$$
\psi^\dagger A\psi\rightarrow
\psi^\dagger S^\dagger AS\psi = \psi^\dagger A\psi
\tag{9}
$$
and the quantity $\psi^\dagger A\gamma^\mu\psi$ transforms as a vector:
$$
\psi^\dagger A\gamma^\mu\psi\rightarrow
\psi^\dagger S^\dagger A\gamma^\mu S\psi
= \psi^\dagger A (S^{-1}\gamma^\mu S)\psi.
\tag{10}
$$
We can think of equations (9)-(10) as the motive for the condition (8). In the representation described above, the familiar choice $A\propto\gamma^0$ satisfies the condition (8), but this depends on the representation.