Under general assumptions, the answer is no.
Supersymmmetry generators belong only to the $\left (1/2, 0 \right ), \left( 0, 1/2\right)$ representation of the homogeneous <PERSON> group (so they can carry only spin $1/2$).
The proof of this fact is contained in the Haag-Lopuszanski-Sohnius theorem.
You can find a good review of the theorem in the final section of this article about Coleman-Mandula theorem: the two are very similar and HSL proof stronly relies on CM theorem.
Now a brief idea of the proof of HSL.
The assumptions of HSL theorem are basically the same in Coleman-Mandula:
* Poincarè invariant theory
* Finite particle type & discrete mass spectrum; only massive particles
* Non trivial $S$ matrix
* Analytic amplitudes as function of momenta in kinematically allowed region
* Generators of symmetries are distribution in momentum space("technical hypothesis")
Coleman-Mandula theorem states that if these conditions hold, the only bosonic symmetry generators (standard Lie algebra operators with commutation relations) are the <PERSON> generators plus internal scalars.
Notice that, stated in another way, CM means that the bosonic symmetry generators trasnform only belong according to the following representations:
* $\left(0, 0 \right)$ (internal scalars)
* $\left( 1/2, 1/2 \right)$ (the $P^{\mu}$ 4-vector)
* $\left(1,0\right) \oplus \left(0, 1 \right)$ (the homogeneous Lorentz group generators $M^{\mu\nu}$)
No other transformation laws are allowed, and this is the crucial fact used in HSL proof.(If one relaxes the massive hypothesis one can get the conformal group generators, but this doesn't invalidate the proof for HSL barring some technicalities)
Why is this useful to prove the properties of supersymmetry generators?
Well, if you now want to study a generic supersymmetry generator $Q$ (anticommuting operators), just take the anticommutator with its conjugate and decompose it with Clebsch-Gordan.
Notice that the anticommutator between two fermionic charges is of bosonic type (think about the sum of two half-integer angular momentum to convince yourself), so it's a generator constrained by the Coleman-Mandula theorem, and thus can only be one of the listed operators above! (we are basically forcing how the anticommutator transforms, and thus we link this to how $Q$ itself can transform)
If $Q\in \left( j, j'\right)$, then ${Q, Q^{\dagger} }$ has a component in $\left(j + j', j + j'\right)$.
By <PERSON>-Mandula one gets that the only possibility is $j = 1/2,\; j' = 0$ and the conjugate, since other $\left (j + j', j+j'\right)$ would give us bosonic generators that aren't <PERSON> generator. This end the proof.
Of course there are technicalities and subtleties I have omitted for the sake of simplicity (you can find them in the links), but the core idea of the original proof is this basically. 364