Homeomorphism and boundary of a complementary component

Question

Let $X\subset \mathbb R^2$ be compact and connected. My question is whether homeomorphisms of $X$ preserve boundaries of complementary components.

More precisely, let $h:X\to X$ be a homeomorphism.

Question 1. If $U$ is a component of $\mathbb R^2\setminus X$ and $B=\partial U$, does there exist a component $V$ of $\mathbb R^2\setminus X$ such that $h[B]=\partial V$?

I am mostly interested in the case when $X$ is locally connected and $1$-dimensional. Under these assumptions, $B$ is a locally connected continuum, and is essentially a simple closed curve.

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EDIT: Question 1 has a negative answer even for topological graphs, as pointed out in the comments. Maybe the better question is:

Question 2. Let $B$ be the union of all boundaries of complementary components of $X$. Then for every homeomorphism $h$ of $X$, $h[B]=B$.

Note that in the case that $X$ is locally connected, $B$ is equivalent to the set of accessible points of $X$ (i.e. points which are arcwise accessible from $\mathbb R^2\setminus X$).

Answer 1

For the modified question, here is a counter-example:

First, note that the Cantor set $K$ is (topologically) homogeneous: the group of homeomorphisms acts transitively. (One way to see this is by observing that the group $\widehat {\mathbb Z}_p$ is homeomorphic to the Cantor set.) Now, let $X$ denote the suspension of $K$; concretely, if $K$ is embedded in $\mathbb R\times \{0\}\subset \mathbb R^2$, then $X$ is the double cone over $K$ from the points $(0,1)$ and $(0,-1)$. Thus, $X$ is planar and every self-homeomorphism of $K$ extends to a self-homeomorphism of $X$. On the other hand, self-homeomorphisms of $K$ do not preserve "boundary" points (since $Homeo(K)$ acts transitively), where the "boundary" $B(K)$ is understood as the union of boundary points of complementary intervals. Hence, $Homeo(X)$ does not send $B(X)$ to $B(X)$ either. Here $B(X)$ is understood as in your question: union of boundaries of complementary components of $X$.

This example, of course, is not locally connected. There are locally connected examples as well, but they are more complicated, limit sets of certain convex-cocompact Kleinian groups. Here is a sketch. Start with a compact surface $S$ of genus $g\ge 1$ and one boundary component. Now, glue four copies of $S$ ($S_1,...,S_4$) along their boundaries. The result is a complex $W$. Let $F_i$ denote the subsurfaces $S_i\cup S_{i+1}$, $i$ is taken mod $4$, in $W$. Let $\pi$ be its fundamental group. One proves that $\pi$ is isomorphic to a convex-cocompact subgroup $\Gamma< PSL(2,\mathbb C)$. Let $X\subset S^2$ denote the limit set of $\Gamma$. It is a Peano continuum. Up to relabelling and conjugation, peripheral subgroups of $\Gamma$ (stabilizers of components of the domain of discontinuity of $\Gamma$) are $\pi_1(F_1)$, ... $\pi_1(F_4)$. But there is an obvious homeomorphism of $W$ which sends $S_1$ to itself and swaps $S_3$ and $S_2$. The corresponding homeomorphism $h$ of the limit set $X$ will not preserve $B(X)$: The image of the limit set of $\pi_1(F_1)$ under $h$ will not be contained in $B(X)$. This is a "reincarnation" of an exotic self-homeomorphism of the bipartite graph with two vertices and four edges embedded in the plane.

Answer 2

The answer to Question 2 is negative: Let $Y$ be a continuum consisting of the segment $[-1,1]\times\{0\}$ to which a sequence of half circles $C_n$ with radius $\frac1n$ lying in the upper halfplane and centered in the origin is attached. The origin is an accesible point of $Y$. Now let $Z$ be almost the same as $Y$, but with each $C_n$ replaced by its symmetric copy in the lower halfplane for $n$ even. Note that $(0,0)$ is not accesible point of $Z$. But there is a homeomorphism of $Y$ and $Z$ fixing the origin.

Now, let $X=Y \cup (Z+(2,0))$, i.e. we shift $Z$ by two units to the right and join it with $Y$ through the point $(1,0)$. There is a self homeomorphism of $X$ moving $Y$ to $Z+(2,0)$ and vice versa. The accesible point $(0,0)$ of $X$ is mapped to the inaccesible point $(2,0)$ of $X$.