$G^0$ denotes the connected component of the identity. It's easy to show that $G^0$ is topologically characteristic in the sense that it is invariant under every continuous automorphism. Here is a more precise formulation of the title question:
Question 1: Let $G$ be a group and let $H$ be a (normal/characteristic) subgroup. Is there a topology on $G$ making it into a topological group with $H=G^0$?
This would mean that $G^0$ is not intrinsically group-theoretic, since any subgroup can be $G^0$ in some arbitrary topology.
Here is a related question. Can every finite set be mapped into $G^0$ by some automorphism?
Question 2: Let $G$ be a topological group. Given $x_1,\ldots,x_n\in G$, is there an automorphism $\sigma:G\rightarrow G$ (not necessarily continuous) such that $\sigma(x_1),\ldots,\sigma(x_n)\in G^0$?
Since continuity of $\sigma$ is not required here, the first question would imply that you can replace $G^0$ by any (normal) subgroup $H$ and always send a finite set into any subgroup $H$. This sounds unlikely: for example, take $H=Z(G)$ to be the center, and $x_i$ any noncentral element. Since $Z(G)$ is preserved by all automorphisms, there's no way that noncentral elements can be mapped into it.
For Q2, we know each $x_i$ is in some coset of $G^0$, where we fix a family $S$ of coset representatives. Write $x_i=g_is_i$ where $g_i\in G^0$ and $s_i\in S$. We can attempt $x_i\mapsto s_i$ but there's no reason to believe that this can be extended to an automorphism of $G$. Is there a relevant group-theoretic property that would help this work?
How would things change if $G$ is an algebraic group instead of a topological group?