  For a fixed graph $H$, the reconfiguration problem for $H$-colourings (i.e.
homomorphisms to $H$) asks: given a graph $G$ and two $H$-colourings $\varphi$
and $\psi$ of $G$, does there exist a sequence $f_0,\dots,f_m$ of
$H$-colourings such that $f_0=\varphi$, $f_m=\psi$ and $f_i(u)f_{i+1}(v)\in
E(H)$ for every $0\leq i<m$ and $uv\in E(G)$? If the graph $G$ is loop-free,
then this is the equivalent to asking whether it possible to transform
$\varphi$ into $\psi$ by changing the colour of one vertex at a time such that
all intermediate mappings are $H$-colourings. In the affirmative, we say that
$\varphi$ reconfigures to $\psi$. Currently, the complexity of deciding whether
an $H$-colouring $\varphi$ reconfigures to an $H$-colouring $\psi$ is only
known when $H$ is a clique, a circular clique, a $C_4$-free graph, or in a few
other cases which are easily derived from these. We show that this problem is
PSPACE-complete when $H$ is an odd wheel.
  An important notion in the study of reconfiguration problems for
$H$-colourings is that of a frozen $H$-colouring; i.e. an $H$-colouring
$\varphi$ such that $\varphi$ does not reconfigure to any $H$-colouring $\psi$
such that $\psi\neq \varphi$. We obtain an explicit dichotomy theorem for the
problem of deciding whether a given graph $G$ admits a frozen $H$-colouring.
The hardness proof involves a reduction from a CSP problem which is shown to be
NP-complete by establishing the non-existence of a certain type of
polymorphism.
