Is every bijection that preserves element order an isomorphism? I know the converse is true, and I suspect the statement is not true, but I can't think of counter example. 
 A: No: take a cyclic group $G$ of prime order $p$. Then all the elements but $0$ have the same order $p$, but it should be clear that you can set up a bijection from $G$ to itself that doesn't do the composition correctly: take $p=5$, for example, and set
$$ f(0)=0, \quad f(1)=1, \quad f(2)=3, \quad \&c. $$
Then $f(1)+f(1)=2 \neq 3= f(2)=f(1+1)$, for example.
A: No.  In fact, it is even possible for two finite groups $G$ and $H$ to have the same number of elements of each order, but still not be isomorphic.  I believe the smallest example of this is $G = \mathbb Z/4 \times \mathbb Z/4$ and $H = \mathbb Z/2 \times Q_8$, where $Q_8$ is the quaternion group.  Each of these has one identity, three elements of order 2, and twelve elements of order 4.  So there exist order-preserving bijections between them, but none of them is an isomorphism, as $G$ is abelian and $H$ is not.
A: Considering $G$ to be any non-abelian group, the map $x\mapsto x^{-1}$ is a bijection, preserving orders of elements, which is not an isomorphism (since .....?......).
