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.

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.

For other cardinals:

Todorcevic´s function and the corresponding graph can be constructed for any successor cardinal so for those the answer is no. For weakly compact cardinals we can repeat the countable case argument so the answer is yes for those. I guess this still leaves infinitely many open questions.