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.