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.