Let $G$ be a finite simple group in which there is no element of order $p^2$ for all primes $p\mid\vert G\vert$. Suppose that $H$ is a finite group whose number of nontrivial proper subgroups is as same as $G$ and there is a bijection $\phi$ from the set of nontrivial subgroups of $G$ to the set of nontrivial subgroups of $H$ such that for every two subgroups of $G$ say $K\_{1}, K\_{2}$ we have $K\_{1}\leq K\_{2}$ if and only if $\phi(K\_{1})\leq \phi(K\_{2})$(i.e. $\phi$ preserves inclusions) 


Can we say that $G\cong H$?



A criterion for determining simple groups
The answer is "yes". Indeed, much more is true. Refer to the following paper:


*Michio Suzuki*, MR 39717 [**On the lattice of subgroups of finite groups**](http://www.ams.org/mathscinet-getitem?mr=39717), *Trans. Amer. Math. Soc.* **70** (1951), 345--371.


The bijection that is mentioned by the OP is a *lattice isomorphism*. In the cited paper, Suzuki proves that if $G$ is simple and is lattice isomorphic to a group $H$, then $H$ is simple. This result does not use the Classification of Finite Simple Groups.


Now, using the Classification of Finite Simple Groups, one can conclude that $G\cong H$. This last statement can be found on p.434 in:



> 
> [*Groups and lattices*](http://www.math.hawaii.edu/~williamdemeo/latticetheory/Palfy-GroupsAndLattices-GStA-2001.pdf), Palfy, P. In the Proceedings of Groups St Andrews 2001.
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> 
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None of these results require the restriction on element orders given in the first sentence of the question.


