The simple connected graph $G$ has $n$ vertices and we have:
1) $E(G)\geq \frac{n(n1)}{3}$
2) we have the spectrum and degree sequence of $G$
3) $Spectrum(G)=Spectrum(H)$
Is $G \cong H$?
The simple connected graph $G$ has $n$ vertices and we have: 1) $E(G)\geq \frac{n(n1)}{3}$ 2) we have the spectrum and degree sequence of $G$ 3) $Spectrum(G)=Spectrum(H)$ Is $G \cong H$? 


No. $H$ is isomorphic to $G$ if and only if the same is true of their complements. The complements of $G$ and $H$ are required to be mildly sparse graphs (certainly $k$regular for any $k$ is fine, as long as $n$ is reasonably large), and one can certainly find $k$ regular isospectral nonisomorphic graphs. EDIT To find such families, google "isospectral cubic graphs". Some of the things which come up are NOT cubic (e.g., the paper Isospectral Cayley graphs of some finite simple groups by Lubotzky, Samuels, Vishne gets examples for almost all degrees.) 


Connected strongly regular graphs with parameters $(v,k,\lambda)$ have eigenvalue $k$ appearing once and two other eigenvalues with prescribed multiplicity. In general there are many nonisomorphic graphs for a fixed parameter. There are two graphs with parameters $(16,6,2)$ and 15 with parameters $(25,12,5)$. The number of non isomorphic graphs can grow dramatically.... All strongly regular graphs with parameter $(v,k,\lambda)$ are regular (so the degree sequence is obvious) and are cospectral. Either the graph or its complement meets your bound on the number of edges. Andries Brouwer maintains a nice webpage on strongly regular graphs here. (Check out the number of $(36,15,6)$ strongly regular graphs!) 

