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?

Let
$\quad \mathcal I(\mathbb N) = \{A \in \mathcal P(\mathbb N) \, | \, A \text{ is infinite} \} $
By the Schröder–Bernstein theorem, if we can construct an injection $\Phi$ from $\mathcal P(\mathbb N)$ into $\mathcal I(\mathbb N)$, then we can claim that $|\mathcal I(\mathbb N)| = \mathcal |\mathcal P(\mathbb N)|$.
Observe that by using this technique it is not necessary to determine beforehand the cardinality of all the finite subsets of $\Bbb N$.
Set $K = \{k \in \mathbb N \mid k \text{ is odd}\}$ and define
$$ \Phi(A) = \begin{cases} \{ 2n \mid n\in A \} & \text{if $A$ is infinite} \\
\{ 2n \mid n\in A \} \cup K& \text{if $A$ is finite}  \end{cases} $$
