How many subsets of $\mathbb{N}$ have the same cardinality as $\mathbb{N}$? 
How many subsets of $\mathbb{N}$ have the same cardinality as $\mathbb{N}$?

I realize that any of the class of functions $f:x\to (n\cdot x)$ gives a bijection between $\mathbb{N}$ and the subset of $\mathbb{N}$ whose members equal multiples of n.  So, we have at least a countable infinity of sets which have the same cardinality of $\mathbb{N}$.  But, we could remove a single element from any countably infinity subset of the natural numbers and we still will end up with a countably infinite subset of $\mathbb{N}$.  So (the reasoning here doesn't seem quite right to me), do there exist uncountably many infinite subsets of $\mathbb{N}$ with the same cardinality of $\mathbb{N}$?  
Also, is the class of all bijections $f: \mathbb{N} \to \mathbb{N}$ and a given countably infinite subset uncountably infinite or countably infinite?

As great answers have been given already, I'd merely like to add an easy way to show that the set of finite subsets of $\mathbb{N}$ is countable:
Observe that
$$\operatorname{Fin}(\mathbb{N}) = \bigcup_{n\in\mathbb{N}}\left\{ A\subseteq\mathbb{N}: \max(A) = n \right\},$$
which is a countable union of finite sets as for each $n\in\mathbb{N}$ there certainly are less than $2^n = \left|\mathcal{P}(\{1,\ldots,n \})\right|$ subsets of $\mathbb{N}$ whose largest element is $n$. Hence, $\operatorname{Fin}(\mathbb{N})$ is countable itself.
From here on, you can use Asaf's argument to show that the set of infinite subsets of $\mathbb{N}$ (which are precisely the sets with the same cardinality as $\mathbb{N}$) must be uncountable.
