The standard example is as follows: Take a $3$-fold $X$; for example let $X=\mathbb{P}^3$. Let $\sigma$ be an automorphism of $X$ of order $n$; for example, $(x_0:x_1:x_2:x_3) \mapsto (x_1:x_2:x_3:x_0)$. Let $C_1$, $C_2$, ..., $C_n$ be an $n$-gon of genus $0$ curves, with $C_i$ meeting $C_{i-1}$ and $C_{i+1}$ transversely and disjoint from the other $C_j$'s, with $\sigma(C_i)=C_{i+1}$. For example, $C_1 = \{(*:*:0:0) \}$, $C_2 = \{(0:*:*:0) \}$, $C_3 = \{(0:0:*:*) \}$ and $C_4 = \{(*:0:0:*) \}$. Let $p_i=C_{i-1} \cap C_{i}$.
Using this input data, we make our example. Take $X \setminus \{ p_1, p_2, \ldots, p_n \}$ and blow up the $C_i$. Call the result $X'$. Also, take a neighborhood $U_i$ of $p_i$, small enough to not contain any other $p_j$. Blow up $C_i \cup cap U_i$, then blow up the proper transform of $C_{i-1}$. C_{i-1} \cap U_i$. Call the result$U'_i$. Glue together$X'$and the$U'_i$'s to make a space$Y$. Clearly,$\sigma$lifts to an action on$Y$. There is a map$f: Y \to X$. The preimage$f^{-1}(p_i)$consists of two genus zero curves,$A_i$and$B_i$, meeting at a node$q_i$. The$q_i$form an orbit for$\sigma$. We claim that there is no affine open$W$containing the$q_i$. Suppose otherwise. The complement of$W$must be a hypersurface, call it$K$. Since$K$does not contain$q_i$, it must meet$A_i$and$B_i$in finitely many points. Since$W$is affine,$W$cannot contain the whole of$A_i$or the whole of$B_i$, so$K$meets$A_i$and$B_i$. This means that the intersection numbers$K \cdot A_i$and$K \cdot B_i$are all positive. We will show that there is no hypersurface$K$with this property. Proof: Let$x$be a point on$C_i$, not equal to$p_i$or$p_{i+1}$. Then$f^{-1}(x)$is a curve in$Y$. As we slide$x$towards$p_i$, that curve splits into$A_i \cup B_i$. As we slide$x$towards$p_{i+1}$, that curves becomes$A_{i+1}$. (Assuming that you chose the same convention I did for which one to call$A$and which to call$B$.) Thus,$[A_i] + [B_i]$is homologous to$A_{i+1}$. So$\sum [A_i] = \sum [A_i] + [B_i]$,$\sum [B_i] = 0$, and$\sum B_i \cdot K = 0$. Thus, it is impossible that all the$B_i \cdot K$are positive. QED I think this example is discussed in one of the appendices to Hartshorne. 1 The standard example is as follows: Take a$3$-fold$X$; for example let$X=\mathbb{P}^3$. Let$\sigma$be an automorphism of$X$of order$n$; for example,$(x_0:x_1:x_2:x_3) \mapsto (x_1:x_2:x_3:x_0)$. Let$C_1$,$C_2$, ...,$C_n$be an$n$-gon of genus$0$curves, with$C_i$meeting$C_{i-1}$and$C_{i+1}$transversely and disjoint from the other$C_j$'s, with$\sigma(C_i)=C_{i+1}$. For example, $C_1 = \{(*:*:0:0) \}$, $C_2 = \{(0:*:*:0) \}$, $C_3 = \{(0:0:*:*) \}$ and $C_4 = \{(*:0:0:*) \}$. Let$p_i=C_{i-1} \cap C_{i}$. Using this input data, we make our example. Take $X \setminus \{ p_1, p_2, \ldots, p_n \}$ and blow up the$C_i$. Call the result$X'$. Also, take a neighborhood$U_i$of$p_i$, small enough to not contain any other$p_j$. Blow up$C_i \cup U_i$, then blow up the proper transform of$C_{i-1}$. Call the result$U'_i$. Glue together$X'$and the$U'_i$'s to make a space$Y$. Clearly,$\sigma$lifts to an action on$Y$. There is a map$f: Y \to X$. The preimage$f^{-1}(p_i)$consists of two genus zero curves,$A_i$and$B_i$, meeting at a node$q_i$. The$q_i$form an orbit for$\sigma$. We claim that there is no affine open$W$containing the$q_i$. Suppose otherwise. The complement of$W$must be a hypersurface, call it$K$. Since$K$does not contain$q_i$, it must meet$A_i$and$B_i$in finitely many points. Since$W$is affine,$W$cannot contain the whole of$A_i$or the whole of$B_i$, so$K$meets$A_i$and$B_i$. This means that the intersection numbers$K \cdot A_i$and$K \cdot B_i$are all positive. We will show that there is no hypersurface$K$with this property. Proof: Let$x$be a point on$C_i$, not equal to$p_i$or$p_{i+1}$. Then$f^{-1}(x)$is a curve in$Y$. As we slide$x$towards$p_i$, that curve splits into$A_i \cup B_i$. As we slide$x$towards$p_{i+1}$, that curves becomes$A_{i+1}$. (Assuming that you chose the same convention I did for which one to call$A$and which to call$B$.) Thus,$[A_i] + [B_i]$is homologous to$A_{i+1}$. So$\sum [A_i] = \sum [A_i] + [B_i]$,$\sum [B_i] = 0$, and$\sum B_i \cdot K = 0$. Thus, it is impossible that all the$B_i \cdot K\$ are positive. QED