In Evasiveness of Graph Properties and Topological Fixed-Point Theorems on page 23 the following is said:
Consider the following two-dimensional complex $\Sigma$ in Figure 2.13, which consists of the sets $\{0, 1, 2\}$, $\{0, 2, 3\}$ and all of their proper non-empty subsets. If we let $f : \Sigma \to \Sigma$ be the simplicial automorphism that sends $\{1 \}$ to $\{3\}$ and leaves $\{0\}$ and $\{2\}$ fixed, then $\Sigma^f$ is a subcomplex of $\Sigma$. However, if we let $h : \Sigma \to \Sigma$ be the simplicial automorphism which transposes $\{ 0 \}$ to $\{ 2 \}$ and leaves $\{ 1 \}$ and $\{ 3 \}$ fixed, then $\Sigma^h$ (? they say $\Delta^h$) is not a subcomplex of $\Sigma$, since it contains $\{ 0, 2 \}$ but does not contain the subset $\{ 0 \}$ and $\{ 2 \}$.
I would have guessed the other way around as switching $\{ 1 \}$ and $\{ 3 \}$ moves them to two different triangles, where as switching $\{ 0 \}$ and $\{ 2 \}$ preserves the triangles of every vertex. I'm not exactly sure how they are "removing" $\{ 0 \}$ and $\{ 2 \}$ as subsets.
Notice that the largest sets are $\{ 0, 1, 2 \}$ and $\{0, 2, 3 \}$. Thus, there is no $\{ 1, 3 \}$ as a proper subset however, there is a $\{ 0, 2 \}$. So: $h(\{0, 2 \}) = \{2, 0 \} = \{0, 2 \}$ and therefore $\{0, 2 \}$ is $h$-invariant but neither $\{ 2 \}$ nor $\{ 0 \}$ but for $\Sigma^h$ to be a simplex, it must contain $\{0\}$ and $\{2\}$. However, this is not an issue for $\Sigma^f$ as $\{1, 3\}$ is not even part of $\Sigma$.