Hmm but in a field you can't divide by zero, so you are demanded a multiplicative group only after you strip zero away. In \$$(\mathbb{Z}/2, \cdot)\$$ you have two actions in \$$\\{0,1\\}\$$: the identity, and the one that zeroes all (and obeys \$$f \circ f = f\$$). You exclude the later, and you get a singleton group where the identity is its own inverse. You have operation tables [here](https://en.wikipedia.org/wiki/GF(2)).

EDIT: to actions -> *two* actions