Let $G$ be a finite group and $A\leq \operatorname{Aut}(G)$. Assume that $P$ is a Sylow $p$-subgroup of $G$ and $z\in Z(P)$ of order $p$ such that each subgroup of order $p$ of $P$ is $A$-conjugate to $\left\langle z\right\rangle $. Can we prove that $N_{GA}(P)$ acts transitively on the set of subgroups of order $p$ of $Z(P)$?
Moreover, can we conclude that $P$ is either homocyclic or generalized quaternion or $2$-automorphic $2$-group?
I think the first part is correct as stated, by the Burnside argument. Let $V=\langle z\rangle,$ and let $W$ be a subgroup of order $p$ of $Z(P).$ Since in particular $W\subseteq P,$ there is $a\in A$ such that $V^a=W.$ Then $V$ centralizes $P$ and $P^{a^{-1}},$ so $P$ and $P^{a^{-1}}$ are Sylow $p$-subgroups of $C_G(V).$ Hence there is $b\in C_G(V)$ with $P^{a^{-1}}=P^b.$ That is, $P=P^{ba}.$ Then $ba\in N_{GA}(P)$ and $V^{ba}=V^a=W,$ as required.
Without further conditions, the answer to the second part of the question is "no" in general. For example, we may take $G = A =Th$ (the sporadic simple group of Thompson) and $p=5$, $P$ a Sylow $p$-subgroup of $G$. Then all subgroups of order $5$ in $P$ are conjugate in $G$ to $Z(P)$, but $P$ is extraspecial of order $125.$
Later edit: Thanks to @TomWilde for the correct definition of $2$-automorphic $2$-group. In the case $p=2$, we can take $G=A$ to be any simple group whose Sylow $2$-subgroup $P$ is dihedral (but not Klein 4). Then all involutions of $G$ are conjugate, but the automorphism group of $P$ is a $2$-group acting trivially on $Z(P)$, so is certainly not transitive on involutions of $P$. Also, $P$ is neither homocyclic nor generalized quaternion.
Even later edit: If the Sylow $p$-subgroup $P$ of $G$ is Abelian, and $A$ has order coprime to $p$, then it is true that $P$ is homocyclic under the hypotheses of the question, because in that case, a Hall $p^{\prime}$-subgroup of $N_{GA}(P)$ acts transitively on subgroups of order $p$ of $P$, so must be acting indecomposably on $P$.
