Cardinality of the set of automorphisms of $(\mathbb{N},+)$ I wonder if the set of bijections $\sigma\,:\mathbb{N}\to\mathbb{N}$ that satisfy
$$
\sigma(a+b) = \sigma(a)+\sigma(b)\qquad \forall a,b\in\mathbb{N}
$$
is countable or uncountable. What if we also ask
$$
\sigma(ab) = \sigma(a)\sigma(b)\qquad\forall a,b\in\mathbb{N}
$$
I know that if we just ask bijection there are uncountable many of them, but i wonder if with these restrictions we get just countable many.
 A: If we use the convention that $0 \in \Bbb N$, we have
$$\sigma(0) = \sigma(0 + 0) = \sigma(0) + \sigma(0),$$
so
$$\sigma(0) = 0.$$
So, for either convention, it's enough to find bijections from $\{1, 2, 3, \ldots\}$ to itself.
Now, we have that $$\sigma(a + 1) = \sigma(a) + \sigma(1) > \sigma(a),$$
that is, that $\sigma$ is strictly increasing. Using that $\sigma$ must be a bijective immediately (or inductively) gives that $$\sigma(n) = n$$
for all $n$, that is, the only such bijection satisfying the criterion is the identity.
Of course, this bijection also satisfies the condition
$$\sigma(ab) = \sigma(a) \sigma(b).$$
A: HINT: First show that $\sigma(1)=1$, then proceed by induction. 
A: There is only one bijection $\sigma: \mathbb N_{> 0} \rightarrow \mathbb N_{> 0}$ s.t. $\sigma(a+b) = \sigma(a)+ \sigma(b)$, namely the identity:
Suppose not, then there is a least $a$ s.t. $\sigma(a) \neq a$. As this is impossible for $a=1$, we have $\sigma(a-1) = a-1$ and $\sigma(1) = 1$ and thus $\sigma(a) = \sigma( (a-1)+1) = \sigma(a-1) + \sigma(1) = a-1 + 1 = a$. (Contradiction)
