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Let's call $\frak{G}$ the class of all groups. Let's consider on $\frak{G}$ the equivalence relation $\sim$ such that $G \sim G' \Leftrightarrow \exists \varphi $ isomorphism of groups such that $G=\varphi(G')$
Consider $\chi: (\frak{G}/\sim) \rightarrow (\frak{G}/\sim)$ such that $\chi([G]_{\sim})=[Aut(G)]_{\sim}$.
What can I say about $\chi$ (one-to-one, onto)? What if I just consider finite groups?
To begin with, there is no such thing as the 'set of all groups.' What you are looking for is a proper class, for which this question should be helpful.
In response to the part about $\chi$ being injective, this is surely not true, as many groups have the same automorphism group - the trivial group and $\mathbb{Z}_2$, for a simple example.
An interesting paper by Iyer contains numerous other examples of groups with the same automorphism groups, in particular Prop. 6.1 - 6.6. He also shows that a nonabelian simple group has the same automorphism group as its covering group.
Furthermore $\chi$ is not surjective. For example, $\mathbb{Z}_p$ is not an automorphism group of any group for $p>2$. (Jack Schmidt says it best here.) Additionally, there is no group $G$ such that $\text{Aut}(G)$ is an Abelian $p$-group of order $\leq p^{11}$ for $p>2$. (reference.) It is also worth noting that Iyer proves that $S_6$ is not the automorphism group of any finite group, and that every finite group occurs as the automorphism group of at most finitely many finite groups.
$\operatorname{Aut}(\Bbb Z/4\Bbb Z)\simeq\operatorname{Aut}(\Bbb Z/3\Bbb Z)$.
There's also another thing which can be said about the map above: it cannot be extended to a functor: this is an exercise from Barr and Wells' Topos Triples and Theories. A prove of this fact is given by the observation that $GL_3(\mathbf F_2)$ the group of automorphisms of $\mathbb F_2^3\cong (\mathbb Z/2 \mathbb Z)^3$ is simple, so the only non injective homomorphism from this group to any other group must be the null homomorphism. We have trivial embedding of $\mathbb F_2^2$ in $\mathbb F_2^3$ and a trivial projection from this to $\mathbb F_2^2$: $$\mathbb F_2^2 \stackrel{i}{\hookrightarrow} \mathbb F_2^3 \stackrel{\pi}{\rightarrow} \mathbb F_2^2 $$
such that $\pi \circ i = 1_{\mathbb F_2^2}$. If there was a functor $\chi$ as above we should get a diagram of type $$\chi(\mathbb F_2^2) \stackrel{\chi{i}}{\rightarrow} \chi(\mathbb F_2^3) \stackrel{\chi{\pi}}{\rightarrow} \chi(\mathbb F_2^2)$$ where $\chi(\pi) \circ \chi (i) = 1_{\chi(\mathbb F_2^2)}$ but $\chi(\pi)$ should be a null homomorphism, because $\chi(\mathbb F_2^3)$ has 168 elements while $\chi(\mathbb F_2^2)$ has just six elements (so the homomorphism cannot be injective). So we get the absurd and it follows that $\chi$ cannot be extended to a functor.
