Can supersymmetries change under dualities, like gauge symmetries can?

Yesn't.

If by duality you mean an exact duality, then yes: dual theories have the exact same amount of supersymmetry. The reason is that two theories are dual if they are actually the same theory, only expressed in different variables. The symmetry of a theory is intrinsic to the theory itself, not to the variables you use to express it (unlike gauge symmetries). In particular, the two theories have the exact same Hilbert space, the same spectrum, same algebra of observables, etc. Whether a theory is supersymmetric can be decided by looking at the Hilbert space, so if the dual theories share the latter, they also share the former.

Nowadays people also use the word duality in a weaker sense, initiated by the work of Seiberg. In this weaker sense, often referred to as an infrared duality, the two theories are actually different, but become identical in some limit, say, at low energies. The two theories describe the same long distance physics, they have "the same vacuum". But the excited states, the short-distance physics, are different.

Weakly dual theories may have different symmetries. The reason is that symmetries may be emergent in the infrared. In this case, one of the theories may have more symmetry than the other, but they both end up with the same symmetry in the limit. Again, see the work of Seiberg for as many examples as one may wish for. His original work was in 4d $$\mathcal N=1$$ supersymmetry, but the cleanest examples occur in 3d, cf. e.g. https://arxiv.org/abs/1702.07035, where they also comment on SUSY enhancement. See also the nice work of Benini and collaborators, e.g. https://arxiv.org/abs/1803.01784. Here you will find several examples of (conjectured) dualities where the different theories have different amounts of SUSY.

You can find many more papers by googling "infrared supersymmetry 3d" or something like that. Fascinating topic. Have fun!

Sketch of the argument.

Take two theories $$T_1,T_2$$, which are assumed to be dual in the strict sense. This means, in particular, that these two theories have the same Hilbert space $$\mathcal H$$ (and the same algebra of observables, such as Poincaré). Assume that $$T_1$$ is supersymmetric, i.e., there exist some operators $$Q_i\in\mathrm{End}(\mathcal H)$$ with the usual super-commutation relations. It is trivial to show that $$T_2$$ also has these operators, i.e., it is also supersymmetric.

The argument is straightforward. In $$T_1$$ all states are paired up in a supersymmetric fashion: to each boson $$|b\rangle\in\mathcal H$$ there is a fermion $$|f\rangle= Q|b\rangle\in\mathcal H$$ (more precisely, there is a whole multiplet whose details depend on the number of spacetime dimensions and amount of supersymmetry $$i=1,2,\dots,\mathcal N$$).

The states $$|b\rangle,|f\rangle$$ also exist in $$T_2$$, by assumption: the two theories have the same Hilbert space and same spectrum of states. This means that, in $$T_2$$, we also have the supercharges. For example, we can define these intrinsically by specifying its matrix elements, $$\langle b|Q|f\rangle:=1$$, etc. It may well be the case that, in $$T_2$$, the supercharges do not act on the fundamental fields (those that appear in the Lagrangian). It would act on non-local operators, or on non-perturbative operators (such as monopoles). In such case, it would be very difficult to guess that $$T_2$$ is also supersymmetric. But it must be, if $$T_1$$ and $$T_2$$ really are dual. If a symmetry exists in $$T_1$$, it must also exist in $$T_2$$, and vice-versa. Sometimes theories are more (super)symmetric than one might initially guess.

If $$T_1$$ and $$T_2$$ are weakly dual, then they do not share all of $$\mathcal H$$, but only the vacuum sector. In such case, the argument above breaks down.

Yes. For starters, $$T$$-duality (is conjectured to) relate string theories with different kinds/numbers of super-symmetries.