How to prove this operation is unique 
Given a prime number $p$, an operation “$*$" is defined over the $N^{+}$ structure of Abel group and satisfies:

*

*$a*b\le a+b$


*All nonzero elements have order $p$
Question: Is this operation unique?

I think this is the case. But I can't explain why.
This problem is from Algebraic problem book exercise by (zhi xu yang),It is said the author choose AMM problem 1978
 A: Here is a start.  It needs completion.  I will assume $p \gt 2$.  Similar arguments will work for $p=2$, but some details need to be cleaned up. 
First note $0*0=0$ by property $1$, so $0$ is of order $1$, so all other elements have order $p$.  Then $0*1=1, 1*1=2$ because otherwise $1$ has a lower period.  Now $1*n=n+1$ for $n \lt p-1, 1*(p-1)=0$ to get the period right.  Now we can show $2*2=4$ and so on.  We can continue to fill in the table and will show that $*$ is exactly addition $\pmod p$ for $a,b \lt p$.  If $p=N$, we are done.
If $N \gt p$, we have $p$ in our set.  We can't have $1*p$ be anything less than $p$ except $2$ because we get a lower period for $1$.  In general, we can show that if $a \lt p, a*p \lt p$, then $a*p=a+a$  I haven't seen how to exclude this.  Once we do, we will get that there must be at least $p^2$ elements by following this route.  I believe eventually we will show that there must be $p^n$ elements and $*$ is addition in the field of $p^n$ elements.   
Added:  If we are given that $*$ is an Abelian group operation we can get there. $1*p$ cannot be anything less than $p+1$ because we can't have $1*p=1*n$ for distinct $n,p$ in a group.  We continue up the same way finding that $*$ is addition $\pmod p$ up to $2p-1$.  As $p$ must have period $p$, we can show that all the elements up to $p^2-1$ are in the group and $*$ is the claimed addition in the field of $p^2$ elements up this high.  Then if $p^2$ is in the group we must have everything up to $p^3-1$ and so on.
