What does the Outer automorphism group act on? If an automorphism preserves conjugacy classes, is it inner? (Let $x^G:=\{gxg^{-1}:g\in G\}$ denote the conjugacy class of $x$)
Am I right in thinking $\text{Out }G := \text{Aut }G/\text{Inn }G$ acts on the conjugacy classes of G, $\{x^G:x\in G\}$? What else could I think of it acting on (trying to get some intuition)?
In addition to that, if $f\in \text{Aut }G$
$$f\in\text{Inn }G\iff(\forall x\in G)f(x^G) = x^G$$
I feel this must be true, but couldn't find anything on it.
If it is true, can we find a weaker condition on the right hand side to check if $f\in\text{Inn }G$? eg. we only require $f(x^G) = x^G$ for $x$ in a set of generators for this to be true.
 A: $\mathrm{Out}G$ acts on the set of conjugacy classes.
Your claim is not true for some infinite groups. An automorphism preserving conjugacy classes is called a conjugating automorphism or class preserving automorphism and the property that every conjugating automorphism is inner is called the Grossman property. See, for example this paper or this question .
For finite groups there is apparently a theorem by Gary Seitz and Walter Feit (1984) that finite simple groups satisfy the Grossman property (I could not find the paper). But in general for finite groups it is not true,
A: Here is a simple example where the Grossman property is violated:
Let $X$ be an infinite set. A permutation $\alpha$ of $X$ is said to have finite support  if there exists a finite subset $F\subset X$ such that $\alpha$ fixes every element of $X-F$.
The set $Aut_f(X)$ of permutations of $X$ with finite support clearly form a normal subgroup of the group $Aut(X)$ of all permutations of $X$.
Suppose that $\alpha$ is any permutation of $X$. Then for every permutation $\beta\in Aut_f(X)$, the permutations $\alpha \beta \alpha^{-1}$ and $\beta$ are conjugate not only in $Aut(X)$, but also in $Aut_f(X)$ (I leave it to you as an exercise to prove this). At the same time, for each $\alpha$ with infinite support, the automorphism of $Aut_f(X)$ defined by
$$
\beta \mapsto \alpha \beta \alpha^{-1}
$$
is not inner (as an automorphism of $Aut_f(X)$). This is another easy exercise.
