jmc

Working from the inside out, we start by computing $G(1)$. Since $(1,-3)$ is on the graph, we have $G(1)=-3$.

Therefore, $G(G(1))=G(-3)$. Since $(-3,5)$ is on the graph, we have $G(G(1))=5$.

Therefore, $G(G(G(1)))=G(5)$. Since $(5,5)$ is on the graph, we have $G(G(G(1)))=5$, and we also see that applying $G$ any number of additional times will leave us at $5$. Therefore, $G(G(G(G(G(1)))))=\boxed{5}$.