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 $C_i \cap U_i$, then blow up the proper transform of $C_{i-1} \cap U_i$. 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.