Since for comment is long, I write the contribution here:

Let $A$ and $L$ be the adjacency matrix and Laplacian matrix of the graph $G$, respectively. Also, suppose $\mu_2$ denotes the second smallest eigenvalue of $L$ and $\lambda_2$ be the second largest eigenvalue of $A$. It is proved that $i(G)\geq \frac{\mu_2}{2}$ (see Theorem 4.1 of *"Isoperimetric numbers of graphs"* by *Bojan Mohar*). So, we have $$i(G)+i(G^c)\geq \frac{\mu_2(G)+\mu_2(G^c)}{2}$$.

So, if we have an example for your question, we must find graph $G$ such that $\mu_2(G)+\mu_2(G^c)<1$. By some spectral graph theory, it can be seen that finding such a graph is a difficult problem. But, finding the spectrum of a graph is much easier than finding its isoperimetric number.

Also, it can be shown that this problem is related to the classification of the graphs based on their second largest eigenvalue of the adjacency matrix. Actually, if we can find a suitable graph such that:
$$\lambda_2(G)+\lambda_2(G^c)\leq \delta(G)-\Delta(G)+n-2$$,

Then we find an example; where $\delta(G)$ and $\Delta(G)$ shows the minimum and maximum degree of the graph $G$ with $n$ vertices.

$\textbf{Added later:}$ By the relation $\mu_2(G)+\mu_2(G^c)<1$ and some random graph testing, it seems that such graphs are so rare. There is not any example of such graph up to $6$ vertices. Also, it seems that we can prove $\mu_2(G)+\mu_2(G^c)\geq 1$, which means there is not such an example.