I have learned recently that $\phi:G \rightarrow S_Y$ is an homomorphism and that it induce a group action, being $G$ any group and $Y$ a nonempty set. Now, I am trying to find the kernel of $\phi$.
My reasoning is that permutation are bijective automorphisms, so its kernel should only contain the identity permutation and if homomorphisms preserve the group structure its kernel should be the same. I can't come up with a counter example or see any flaws in my logic (go figure, right?), is this statement correct?
Regards
Note that permutations of $Y$ are just bijective maps (not automorphisms), as $Y$ is a pure set. Hence it does not make sense to speak of their kernel. On the other hand, even if $Y$ were a group, the kernel of the homomorphism $\phi$ has nothing in common with some kernels of elements of $S_Y$.
What you have to do here, is expressing the kernel of the homorphism $\phi\colon G \to S_Y$ in terms of the associated group action $\Phi \colon G \times Y \to Y$, $(g, y) \mapsto \def\action#1{{}^{#1}}\action g y := \phi(g)(y)$. As you write correctly, the unit of $S_Y$ is the identity map, hence \begin{align*} g \in \ker \phi &\iff \phi(g) = \mathrm{id}_Y \\ &\iff \forall y \in Y: \phi(g)(y) = y\\ &\iff \forall y \in Y: \action g y = y\\ &\iff g \in \bigcap_{y \in Y} \{h \in G: \action h y = y \} \end{align*} Hence, the kernel of $\phi$ is the intersection of the stabilizer groups of all $y \in Y$.