Action of the normalizer of the complement 
Let $G$ be a finite group, and let $p$ be a prime. Let $U$ be a Sylow $p$-subgroup of $G$. By the Schur-Zassenhaus theorem, there is a complement $T$ to $U$ in $N_G(U)$. Observe that $N_G(T)$ acts by conjugation on the set of Sylow $p$-subgroups that are normalized by $T$ (one of which is $U$). Use more Schur-Zassenhaus to show that this action is transitive.

Attempt: Let $X$ be the set of Sylow $p$-subgroups that are normalized by $T$, i.e., for any $U'\in X$, $T\subset N_G(U')$. Then $N_G(T)\subset N_G(U')$. Since Sylow $p$-subgroups are conjugate, suppose $U'\in X$, then $U'=g^{-1}Ug$ for some $g\in N_G(T)$. Then since $g\in N_G(U)$, $g=ut$ for some $t\in T, u\in U$, then $U'=g^{-1}Ug=(ut)^{-1}U(ut)=tUt^{-1}$. Therefore, every element in $X$ is conjugate to $U$ by an element of $T$.
Questions: I feel what I got is not what the question wanted to ask, because I proved the action of $N_G(T)$ on $X$ is in fact action of $T$? Plus I didn't use the SZ theorem. Moreover, I can't see the observation in the problem statement: why does $N_G(T)$ act on the set by conjugation?
 A: There is some confusion in your attempt. First you claim that $T \subset N_G(U')$ implies $N_G(T) \subset N_G(U')$, but this is false. After this you claim that $U$ is conjugate to $U'$ by an element of $N_G(T)$, but this is exactly what you are trying to prove.
To get started with, we have an action of $G$ on the $p$-Sylows by conjugation. Now $N_G(T)$ normalizes $T$, so it acts on the set $X$ of fixed points of $T$ on this action.

In general: If $A$ acts on a set $\Omega$ and $B \trianglelefteq A$, then $A$ acts on the set $$\Omega^B = \{ \alpha \in \Omega : b \cdot \alpha = \alpha \text{ for all } b \in B\}$$ (Fixed points of $B$).

So that is why $N_G(T)$ acts on the set of $p$-Sylows normalized by $T$.
For the actual question, here is a pointer. Take some $p$-Sylow normalized by $T$, say $U' = U^g$. We have $N_G(U) = U \rtimes T$ so $N_G(U^g) = U^g \rtimes T$. Now observe that $T^{g^{-1}}$ is another complement to $U$ in $N_G(U)$ and use the fact that by Schur-Zassenhaus complements are conjugate.
