The defining relation for the Clifford algebra, $Cl(1,d)$ is
$$
\{\gamma_\mu,\gamma_\nu\}=2 \eta_{\mu\nu}\ \mathbf{1}\ ,
$$
For simplicity, I will assume that $\eta_{\mu\nu}=\text{Diag}(1,-1,\ldots,-1)$ with $\mu,\nu=0,1,\ldots,d$. Other signatures can easily be incorporated. It is easy to see that $\gamma_0^2=-\gamma_i^2=\mathbf{1}$ for $i=1,\ldots,d$. Using the defining relation, one has
$$
\gamma_0 \gamma_i + \gamma_i \gamma_0 =0 \ .
$$
Multiply the above equation by $\gamma_0$ and then take the trace to obtain
$$
\text{Tr}(\gamma_i) + \text{Tr}(\gamma_0 \gamma_i \gamma_0)=0\implies \text{Tr}(\gamma_i)=0\ ,
$$
on using the cyclic property of the trace. Similarly, one can show $\text{Tr}(\gamma_0)=0$. So the defining property proves the tracelessness of the Dirac matrices.
Two representations, $\gamma_\mu$ and $\gamma_\mu'$, of the Clifford algebra are said to be equivalent if $\gamma_\mu' = S \cdot \gamma_\mu S^{-1}$ for some invertible matrix $S$.
Appendix A of the Physics Reports article by Sohnius might be a good starting point for the other properties.
Okay, this wasn't as hard to find an answer to as I expected. However, any clarifications/ critisisms are welcome.
Weinberg basically gives the answer in section 5.6 of his QFT book:
A general tensor of rank N transforms as the direct product of N (1/2, 1/2)
four-vector representations. It can therefore be decomposed (by suitable
symmetrizations and antisymmetrizations and extracting traces) into irreducible terms (A,B) with A = N/2, N/2-1,... and В = N/2, N/2-1,... . In this way, we can construct any irreducible representation (A,B) for
which A + В is an integer. The spin representations, for which A + В is
half an odd integer, can similarly be constructed from the direct product
of these tensor representations and the Dirac representation $(1/2,0)\oplus(0,1/2)$.
So it does not appear that there is any simple way to unify the representations for spinor and vector generators, but one can construct the generator for arbitrary half-spins: it has as many copies of $M$ as needed, plus one copy of $S$ if it is a half-integer.
However, the converse is not true. That is, a spin-two field must transform with two copies of $M$, but doing so does not guarantee that an object is spin-two. A counterexample is the electromagnetic tensor $F$, which is certainly spin-one. The difference lies in the ability to equate the two generators as a result of the symmetry properties of the tensor, as elaborated in this answer.
Applying this to the spin 3-2 field, we expect it to have a rotation generator that looks schematically like $S \otimes I+ I \otimes M$. And indeed this is the case- the equivalent of the Dirac equation for spin 3/2 is the Rarita-Schwinger equations:
$\gamma_a \psi^{a}_\mu=0$,
$(i \gamma^\rho \partial_\rho-m)\psi^{a}_\mu=0 $
Which transforms as $\psi'^b _\nu=(\Lambda_\nu^\mu \otimes T^b_a) \psi_\mu^a$,
whose generators are the above combination of $M$ and $S$.
Best Answer
If you know the change of the (vector) basis, the answer is straightforward.
If you don't know the change of the (vector) basis, but only want some particular representation for the gamma matrices (for instance you want only real matrices, or only imaginary matrices), you may try for $S$ :
$$S=\frac{1}{\sqrt{2}} \begin{pmatrix} A&B\\-\epsilon B&\epsilon A \end{pmatrix}, S^{-1}=\frac{1}{\sqrt{2}}\begin{pmatrix} A&-\epsilon B\\ B&\epsilon A \end{pmatrix}$$
where $\epsilon = \pm1$, $A, B$ are $2*2$ matrices, such as $A^2= B^2=1$, and $[A, B]=0$.
For instance, you may take one of the matrix equals to $\pm \mathbb{Id}$, and the other being a Pauli matrix $\pm \sigma_i$.
For obtaining Majorana representation from Dirac representation, we may use : $\epsilon = -1, A = \mathbb{Id}, B = \sigma_y$