I take a nilpotent injector of a finite solvable group $G$ to be a nilpotent subgroup $M$ of $G$ such that $M \cap N$ is a maximal nilpotent normal subgroup of $N$ whenever $N$ is subnormal in $G$. Assuming existence of $M$ , I think uniqueness up to conjugacy follows inductively. Notice that $M \cap H$ is a nilpotent injector of $H$ whenever $H$ is normal in $G.$
We may suppose that $Z(G) = 1$. Now let $p$ be a prime divisor of $|F(G)|$. Since $F(G) \leq M$, we have $O_{p'}(M) \leq C_{G}(O_{p}(G)).$ Thus $O_{p'}(M)$ O_{p'}(M) = O_{p'}(L)$, where $L = M \cap C_{G}(O_{p}(G))$ is a nilpotent injector of $C_{G}(O_{p}(G))$. For notice that $O_{p'}(L) \lhd M$ so that $O_{p'}(L) \leq O_{p'}(M)$, while$O_{p'}(M) \leq M \cap C_{G}(O_{p}(G)) =L$and $O_{p'}(M) \leq O_{p'}(L)$. Now$L$is unique up to conjugacy within $C_{G}(O_{p}(G))$, so certainly unique up to conjugacy in$G$. Hence $O_{p'}(M)= O_{p'}(L)$ is unique up to conjugacy within$G$. By maximality as a nilpotent subgroup, $ M = P \times O_{p'}(M)$, where$P$is a Sylow$p$-subgroup of $C_{G}(O_{p'}(M)).$ Hence we see that$M$is unique up to conjugacy. in$G$. 2 expanded text to clarify answer. I take a nilpotent injector of a finite solvable group$G$to be a nilpotent subgroup$M$of$G$such that$M \cap N$is a maximal nilpotent normal subgroup of$N$whenevr whenever$N$is subnormal in$G$. Assuming existence of$M$, I think uniqueness up to conjugacy follows inductively. Notice that$M \cap H$is a nilpotent injector of$H$whenever$H$is normal in$G.$We may suppose that$Z(G) = 1$. Now let$p$be a prime divisor of$|F(G)|$. Since$F(G) \leq M$, we have $O_{p'}(M) \leq C_{G}(O_{p}(G)).$ Thus $O_{p'}(M)$ is$= O_{p'}(L)$, where $L = M \cap C_{G}(O_{p}(G)$C_{G}(O_{p}(G))$ is a nilpotent injector of $C_{G}(O_{p}(G))$. For notice that $O_{p'}(L) \lhd M$ so that $O_{p'}(L) \leq O_{p'}(M)$, while $O_{p'}(M) \leq M \cap C_{G}(O_{p}(G)) =L$ and $O_{p'}(M) \leq O_{p'}(L)$.
Now $L$ is unique up to conjugacy within $C_{G}(O_{p}(G))$, so certainly unique up to conjugacy in $O_{p'}(M)$G$. Hence $O_{p'}(M)= O_{p'}(L)$ is unique up to conjugacy . Since within$G$. By maximality as a nilpotent subgroup, $ M = P \times O_{p'}(M)$, where$P$is a Sylow$p$-subgroup of $C_{G}(O_{p'}(M))$, C_{G}(O_{p'}(M)).$ Hence we see that $M$ is unique up to conjugacy. in $G$.
I take a nilpotent injector of a finite solvable group $G$ to be a nilpotent subgroup $M$ such that $M \cap N$ is a maximal nilpotent normal subgroup of $N$ whenevr $N$ is subnormal in $G$. Assuming existence of $M$ , I think uniqueness up to conjugacy follows inductively. We may suppose that $Z(G) = 1$. Now let $p$ be a prime divisor of $|F(G)|$. Since $F(G) \leq M$, we have $O_{p'}(M) \leq C_{G}(O_{p}(G)).$ Thus $O_{p'}(M)$ is $O_{p'}(L)$ where $L = M \cap C_{G}(O_{p}(G)$ is a nilpotent injector of $C_{G}(O_{p}(G))$. Now $L$ is unique up to conjugacy, so $O_{p'}(M)$ is unique up to conjugacy . Since $M = P \times O_{p'}(M)$, where $P$ is a Sylow $p$-subgroup of $C_{G}(O_{p'}(M))$, we see that $M$ is unique up to conjugacy.