The answer is yes for countable graphs:
Fix an infinite graph $G$ and a bijective homomorphism $f:G \to G$. Define $c:[G]^2 \to 2$ as $c(\alpha,\beta)=1$ if $\{f\alpha, f\beta\} \in E(G)$ and $c(\alpha,\beta)=0$ otherwise. Since $G$ is infinite, by Ramsey´s Theorem there is an infinite $c$-homogeneous $H \subseteq G$. If $c \upharpoonright [H]^2$ is constant $0$ then $f \upharpoonright H$ is an isomorphism. If $c \upharpoonright [H]^2$ is constant $1$ then $f \upharpoonright f(H)$ is an isomorphism.
The answer is no for graphs of size $\aleph_1$:
Fix a coloring $c:[\omega_1]^2 \to \mathbb{Z}$ with the property that for all uncountable $A\subseteq \omega_1$ and for all $n \in \mathbb{Z}$ there are $\alpha,\beta \in A$ such that $c(\alpha,\beta)=n$. Such a coloring was constructed by S. Todorcevic in the 1980's, using his method of minimal walks.
Define a graph $G$ by $V(G)=\omega_1 \times \mathbb{Z}$ and $E(G)=\{\{(\alpha,n),(\beta,n)\} : c(\alpha,\beta) < n\}$. Consider the function $f:G \to G$ defined by $f(\alpha,n)=(\alpha,n+1)$. It should be clear that $f$ is a bijective homomorphism. However, if $H \subseteq G$ is uncountable, then there is an $n_0 \in \mathbb{Z}$ such that the set $A=\{\alpha: (\alpha,n_0) \in H\}$ is uncountable. So we can find $\alpha, \beta \in A$ such that $c(\alpha,\beta)=n_0$ and thus the pair $\{(\alpha,n_0),(\beta,n_0)\}$ witnesses the fact that $f \upharpoonright H$ is not an isomorphism.