If you equip two isomorphic groups with homeomorphic topologies, are they isomorphic as topological groups? I'm wondering if anyone has any insight regarding the truth of the above statement. Intuitively, if I have two topological groups in which their algebraic group structures are the same up to relabelling, and toplogical spaces that behave the same, it seems as though as topological groups they would have the same structure and behaviours, up to relabelling of course. Or is there an obvious counterexample that I'm missing? 
I've seen a similar question posted with a counterexample, however I believe that the counterexample proposed did not accurately satisfy the hypothesis. 

If $G$ is a finite topological group and $N$ is the connected component of the identity, then $N$ is normal and the coset space $G/N$ forms a basis for the topology. Conversely, one can create any finite topological group given a choice of finite group $G$ and normal subgroup $N$ to be the connected component. (See this question for the result.)
Thus, if we can find a finite group $G$ with two normal subgroups $N_1$ and $N_2$ that are not related by any automorphism but are nonetheless the same size, we can have $(G,\tau_1)$ and $(G,\tau_2)$ homeomorphic, but a continuous isomorphism would have to preserve identity's connected component, i.e. send $N_1$ to $N_2$, which is impossible, and we would have a counterexample.
For this, we can pick $N_1$ and $N_2$ to simply be nonisomorphic. For instance, if $G=\mathbb{Z}_2\times\mathbb{Z}_4$ then we can use the subgroups $N_1=\mathbb{Z}_2\times\mathbb{Z}_2$ and $N_2=\mathbb{Z}_4$.
