Let $\Gamma = (V,E)$ be a vertex-transitive finite graph with $|V| \geq 2$ and $G := \operatorname{Aut}(\Gamma)$. Then there exists $1_G \not= g \in G$ such that $x^g \not= x$ for all $x \in V$. So there is an automorphism of $\Gamma$ without any fixed vertex.
Let's assume that every $g \in G$ has a fixed vertex $v_g \in V$ such that $(v_g)^g = v_g$. Since $V$ has at least two elements, there exists an $w \in V$ with $w^h = v_g$ for an $h \in G$ because $\Gamma$ is vertex-transitive.
Now I used every condition of the claim but I still don't see where to get to.