Fix a positive integer $m$. Let $G = \lbrace h\in\mathbb C | h^m = 1\rbrace$ and $(X,\pi)$ the standard representation of $G$. Namely $X = \mathbb C$ and $\pi:G \to GL(X)$ is defined by $\pi(h)v=h v$ for $h \in G$ and $v \in X$.
Choose $h \in G$ with $h \ne 1$. The fixed point set $X^h$ of the action of $h$ consists only of $0$. Thus $K^G(X^h)$ is isomorphic to the representation ring of $G$. Namely $K^G(X^h) = \mathbb Z[\xi]/(\xi^m-1)$. Here $\xi$ corresponds to the standard representation of $G$.
Let $N_h$ be the normal bundle of $X_h$ in $X$ and $N_h^*$ the dual bundle. Then $[N_h]=\xi$ and therefore $[N_h^*]=[N_h]^{-1}=\xi^{m-1}$ (because $N_h \otimes N_h^* = \mathbb C$).
Define $\lambda_{-1}(N_h^*) = \sum_i(-1)^i[\Lambda^iN_h^*] = 1-\xi^{m-1}$.
According to Thomason (Lem 3.3), $\lambda_{-1}(N_h^*)$ is invertible in $K^G(X^h)$.
But $(1-\xi^{m-1})(1+\xi+\xi^2+\dots+\xi^{m-1})=0$. Therefore $\lambda_{-1}(N_h^*)$ is not invertible.
Can you tell me what is wring in the above discussion?
Note: What I am really reading is Edidin-Graham (§2.3. Just before Prop 2.4.).