Let's say we have $5$ Sylow $p$-subgroups in $G$ for some $p$. This induces a group homomorphism $f: G\to S_{5}$ if we let $G$ act on the set $Syl_p$ by conjugation.
Question: I'm not 100% sure if I understand why it follows that $|f(G)|\geq 5$ from the fact that the group action is transitive.
I would say that we have at least permutations $\sigma$ with $\sigma (i)=j$ for each $i,j\in\{1,\dots,5\}$. So for example, if we fix $i=1$ we get 5 permutations for this that cannot be the same since they are bijections. Is that correct? Maybe someone else has more insight and would care sharing his point of view?
Just use the class formula: there results from this formula that, for any group $G$ operating on a set $X$, the cardinal of an orbit is a divisor of $|G|$.