To see that the number of equivalence relations on an infinite set $A$ of size $\kappa$ is $2^{\kappa}=|{\mathcal P}(A)|$ (i.e., the largest possible size), recall that any set of size $\kappa$ can be split into $\kappa$ sets of size $\kappa$: $\kappa=\kappa\times\kappa$. Say $A$ has size $\kappa$. Fix a partition $A=\bigcup_{i\in I}A_i$, where $I$ is an index set of size $\kappa$, each $A_i$ has size $\kappa$, and $A_i\cap A_j=\emptyset$ if $i\ne j$. Also, for each $A_i$, find a partition $A_i=B_i\cup C_i$ where each $B_i,C_i$ is non-empty.
Given $D\subseteq I$, let $E_D$ be the equivalence relation on $A$ defined as follows:
If $i\in D$, then $B_i$ and $C_i$ are equivalence classes. If $i\in I$ is not in $D$, then the whole of $A_i$ is an equivalence class.
More precisely, in case the above description is not clear: $a E_D b$ for $a,b\in A$, iff both $a,b$ are in the same $A_i$ (for some -unique- $i\in I$) and (if $i\notin D$, then both $a,b$ are in $B_i$ or both are in $C_i$).
Then, from $E_D$, we can reconstruct $D$, so the map $f:{\mathcal P}(I)\to {\mathcal E}$, where ${\mathcal E}$ is the set of equivalence relations on $A$, given by $f(D)=E_D$ is injective, and $|{\mathcal E}|\ge 2^\kappa$.
(In fact, for this inequality to hold, all we needed is that $2\times\kappa=\kappa$: We could have taken each $A_i$ of size 2 and each $B_i,C_i$ a singleton. However, the argument below uses that $\kappa\times\kappa=\kappa$.)
Since each class is a subset of $A\times A$, and different classes are disjoint, each equivalence relation has at most $\kappa=|A|$ classes (cannot have more classes than elements of $A$), and each class is chosen from ${\mathcal P}(A\times A)$, so there are at most $\kappa\times 2^{|\kappa\times\kappa|}=2^\kappa$ equivalence relations.
It follows that the number of equivalence relations is $2^\kappa$, as claimed.
Since each equivalence relation is symmetric, this also shows that there are precisely $2^\kappa$ symmetric relations on $A$ if $|A|=\kappa$.
Let me close with a technical remark: If $A$ is countable (so $\kappa$ is usually denoted $\aleph_0=|{\mathbb N}|=|{\mathbb Z}|$), the equality $\kappa=\kappa\times\kappa$ and the equality $\kappa 2^\kappa=2^\kappa$ can be verified without using the axiom of choice. Same if $A$ has the size of the reals (so $\kappa$ is usually denoted ${\mathfrak c}$ or $2^{\aleph_0}=|{\mathcal P}({\mathbb N})|$).
However, for arbitrary infinite $A$, the equalities require the axiom of choice. I haven't checked what the number of equivalence relations for an arbitrary infinite $A$ is if choice fails. I do not think it is $2^{|A|}$ in general.
Part $1$: The cardinality is uncountable, with a caveat:
The cardinality of this equivalence class is, at first glance, uncountable. For notation's sake, let us denote this class by $\overline{\aleph_0}$, to avoid confusion with the cardinal.
It is known that the cardinality of the reals - that of the continuum - is greater than that of $\aleph_0$ (well-known from Cantor's diagonal argument). Taking this and that $| \Bbb N | = \aleph_0$ as given, consider the sets $\Bbb N \cup \{x\}$ where $x \in \Bbb R \setminus \Bbb N$ is fixed. It is trivial to show that this new set has cardinality equal to that of $\aleph_0$.
Consider the set of all such sets, for variable $x$ a real number but not a natural. It should be clear that since you have a unique set for each real number, then the set is bijective with the reals minus the naturals (which is obviously still the same as the reals in terms of cardinality). You could define the bijection explicitly:
$$f : \Bbb R \setminus \Bbb N \to \bigcup_{x \in \Bbb R \setminus N} \left\{ \Bbb N \cup \{x\} \right\} \;\;\;\;\; x \mapsto \left\{ N \cup \{x\} \right\}$$
(Anecdote, this motivated the unnecessary restriction that $x$ is not a natural number, the simplicity and clearness that $f$ is a bijection. Just in case you thought that was weird.)
Thus, since the sets are bijective, they share cardinality, i.e.
$$\overline{\aleph_0} \geq \left| \bigcup_{x \in \Bbb R \setminus N} \left\{ \Bbb N \cup \{x\} \right\} \right| = |\Bbb R \setminus \Bbb N| = | \Bbb R | = c > \aleph_0$$
Thus, by construction, we have found a subset of $\overline{\aleph_0}$ which is uncountably infinite.
(Note that this does not mean that $|\overline{\aleph_0}| = c$, because it's a subset, and that need not hold. It doesn't make an assertion as to the specific)
Part $2$: The cardinality of $\overline{\aleph_0}$ is actually not well-defined:
The logical follow-up question is obvious: what is the cardinality of $\overline{\aleph_0}$ if not $\aleph_0$?
In fact, there can be no such cardinality assigned to the class.
Notice that our method above essentially allows us to construct a subset of $\overline{\aleph_0}$ of any cardinality. Let $S$ be a set of cardinality $|S|$. Then generalize our argument from before by just letting $x$ be from, not $\Bbb R \setminus \Bbb N$, but $S$ (minus the members of $\Bbb N$ in $S$ if you so choose). Redefining $f$ to account for the obvious changes in the previous argument then shows that $\overline{\aleph_0} \geq |S|$, contradicting the assignment of the cardinal for appropriately chosen $S$.
It is well known that there exists no "maximum cardinal" - that is, there is always a set with a bigger cardinality (always a bigger fish, as it were). You can't assign a value to $\overline{\aleph_0}$, because you have a set with a larger cardinality, from which you could construct subsets of $\overline{\aleph_0}$, and thus have a subset with greater cardinality than $\overline{\aleph_0}$.
It has the sort of flavor of Russel's paradox.
Part $3$: A conclusion:
The conclusion is - to talk about the cardinality of the set $\overline{\aleph_0}$ is mostly meaningless. If it existed, it would be uncountable - but to assign such a cardinality would be contradictory. $\overline{\aleph_0}$ is just that big.
Credits to bof from the comments of the question body for the idea which led to the construction of this argument.
Best Answer
Your approach is sound. I think you can choose a better injection from $P(\mathbb{N})$ into equivalence relations. Let us say $0 \in \Bbb N$ and we will inject $P(\Bbb N \setminus\{0\})$ into the equivalence relations on $\Bbb N$. I would suggest you take a subset of $\Bbb N \setminus\{0\}$ to the equivalence relation that groups all members of the subset and $0$ into an equivalence class and leaves all the rest of $\Bbb N$ under the identity.
If we don't do the trick with $0$, all singleton subsets will be mapped to the identity relation and you don't have an injection.