I am asked to show that if $F$ is a function class and an automorphism of $\left (V,\in \right )$ then F must be the identity, where $V$ is the universe of all sets.
My thoughts currently are that this means if $x,y \in V $ satisfy $x \in y$ and we denote by $F(x)$ the image of $x$ under $F$, then $F(x) \in F(y)$.
Given this interpretation, then we see that $F(\phi) = \phi$, but I think that since $\phi$ doesn't have a non-empty subset, this observation isn't too helpful here.
By trichotomy $F$ either has only fixed points, only one point that is not fixed, or at least two points that are not fixed. If we can eliminate the latter two options, then we are done.
Suppose then than $F(x) \neq x$ and $\forall y \neq x, F(y) = y$
Then $\forall z \in x, F(z) = z \in F(x) \Rightarrow x \subset F(x)$
$F(x) \neq x \Rightarrow \exists y \in F(x) \backslash x, y \neq \phi$
Now note if $x \in F(x)$ then $F(x) \in F(x)$, which I think violate the definition of a set. Hence our $y\neq x$
Finally since $F$ is an isomorphism, we can go backwards from $y \in F(x)$ to get $y \in x $, but by definition $y \notin x.$ So the second case is not possible.
Now suppose we there are at least two elements that are not fixed under $F$. There are two cases to consider here:
$\exists x,y \in V, x \neq F(x), y \neq F(y)$ and $x \in y$ or $\nexists x,y \in V, x \neq F(x), y \neq F(y)$ and $x \in y$
In the latter case, we can repeat our earlier work to reach a similar contradiction.
Suppose then we have elements $x,y \in V$ satisfying $x\in y, x \neq F(x), y \neq F(y)$.
This is where I get stuck. I don't really know how to proceed in this case. I also don't know if I've made a mistake in my working yet, something about it feels off but I'm not sure what.
The observation that $\emptyset=F(\emptyset)$ is in fact the starting point. The other key observation is:
This is proved in two steps. If $x\in y$ then we must have $F(x)\in F(y)$; but $x=F(x)$, so $x\in F(y)$. This shows $y\subseteq F(y)$. The converse ($F(y)\subseteq y$) is shown similarly.
Now combining these, we argue inductively. Remember that every set has a rank - we let $rk(z)$ be the least $\alpha$ such that $z\in V_{\alpha}$. (Sometimes this is defined as the least $\alpha$ such that $z\in V_{\alpha+1}$; it really doesn't matter here.) You now need to argue by induction on rank:
Every set of rank $1$ is fixed by $F$: the only set of rank $1$ is $\emptyset$, and you've already shown that $F(\emptyset)=\emptyset$.
Now suppose that every set of rank $<\alpha$ is fixed by $F$, and $x$ is a set of rank $\alpha$. Can you show that $F(x)=x$? This will show that every set of rank $\alpha$ is also fixed by $F$, and this will complete the induction. (HINT: what can you say about the rank of an element of a set of rank $\alpha$?)