Show that there is no asymmetric simple graph on n vertices.
Note that an asymmetrical graph is one in which no two vertices are similar. (I misread the question -- the above isn't actually true but I wrote a proof and can't understand why it is wrong).
My inductive argument is:
For $n=2$ it is trivial. Now assume the above holds for $n−1$ vertices, and lets consider the case with $n$ vertices. Choose any vertex $v_i$, and let $$E'=\{e\mid e\in E(G), \ e=e(v_i,v)\}$$
Now for the graph with vertex set $V\setminus\{v_i\}$ and edge set $E\setminus E'$, we have that there exists a nontrivial automorphism $g=(\theta,\varphi)$. Let $g'= (\theta',\varphi')$ be the modified automorphism with $\theta'(v)=\theta(v)$ for $v\in V\setminus\{v_i\}$ and $\varphi'(e)=\varphi(e)$ for $e\in E\setminus E'$.
Now let $\theta'(v_i) =v_i$, and let $\varphi'(e)$ be s.t. $\varphi'(e)=e(\theta'(v_i),\theta'(v)) =e(v_i,\theta(v))$ for $e\in E'$. This is a nontrivial automorphism and thus $G$ is not asymmetrical.
Let's start with the $6$-vertex asymmetric graph $G$ below, and see where your proof that it doesn't exist fails.
If we remove vertex $6$, then $G' = G - \{6\}$ is a path on vertices $1, 2, 3, 4, 5$ in that order, and has a nontrivial automorphism given by $$\theta'(1)=5,\; \theta'(2)=4,\; \theta'(3)=3,\; \theta'(4)=2,\; \theta'(5)=1$$ and the corresponding permutation $\phi'$ of the edges. Now, let's try to extend this to an automorphism of $G$ with $\theta(v) = \theta'(v)$ for $v = 1,2,3,4,5$ and $\theta(6)=6$. Your proof says that for an edge $e$ between $v \in \{1,2,3,4,5\}$ and $6$, $\phi(e)$ should map it to an edge between $\theta(v)$ and $6$.
So you're asking $\phi$ to map the edge between $4$ and $6$ to the edge between $2$ and $6$. But that edge doesn't exist! Therefore $(\theta, \phi)$ is not an automorphism of $G$, because $\phi$ sends some of the edges of $G$ to things that aren't edges.