We claim that all such functions are of the form $f(n) = n + c$ for some natural number $c$. The verification is trivial; now consider some function $f$ satisfying the given condition. For $n \in \mathbb{N}$, let $d(n)$ denote the number of 1's in the binary representation of $n$. Let $P(x, y)$ denote the statement that $f(x) + y$ and $f(y) + x$ have the same number of 1's in their binary representation.

Claim 1 For any $y, n \in \mathbb{N}$ with $2^n > f(y)$, $f(2^n - f(y)) + y$ is a power of two.

Proof. $P(2^n - f(y), y)$ gives us that $2^n$ and $f(2^n - f(y)) + y$ have the same number of 1's, and the former has exactly one 1, so $f(2^n - f(y)) + y$ has exactly one 1, from which the claim follows. $\square$

Claim 2 $f(y + 2^k) - f(y)$ is a power of two for any $k \ge 0$ and $y \ge 2^k$.

Proof. Choose an $n$ such that $n > 1000 + \log_2(10 + |f(y + 2^k) - f(y)|)$. By Claim 1, $f(2^n - f(y)) = 2^t - y > 0$ for some $t$ with $t \ge k + 1$. Therefore $P(2^n - f(y), y + 2^k)$ gives
$$d(2^n - f(y) + f(y + 2^k)) = d(2^t + 2^k) = 2$$
since $t \ge k + 1$. If $f(y) = f(y + 2^k)$, then LHS is $d(2^n) = 1$, contradiction! If $f(y) > f(y + 2^k)$, and $f(y) - f(y + 2^k)$ has $m < \log_2(10 + |f(y + 2^k) - f(y)|)$ digits, then
$$d(2^n - f(y) + f(y + 2^k)) \ge n - m - 1 \ge 999 > 2$$
since $2^n - f(y) + f(y + 2^k)$ starts with at least $n-m-1$ ones, contradiction! Therefore $f(y+2^k) > f(y)$, and since $n$ is bigger than the number of digits in $f(y + 2^k) - f(y)$, there is no carry-over, so
$$2 = d(2^n - f(y) + f(y + 2^k)) = 1 + d(f(y + 2^k) - f(y))$$
which gives us $f(y + 2^k) - f(y)$ is a power of 2, as required $\square$

Claim 2 gives us $f(y+1) - f(y) = 2^{t(y)}$ for some $t(y)$, for all $y$. But for $y \ge 2$, $f(y+2) - f(y)$ is also a power of two $\implies 2^{t(y)} + 2^{t(y+1)}$ is a power of two, which is only possible if $t(y) = t(y+1)$ for all $y \ge 2$. Therefore $f(y+1) - f(y)$ is a constant power of two for all $y \ge 2$, say $2^k$. This gives us $f(y) = 2^k y + c$ for some constant $c$, for all $y \ge 2$. Putting this in Claim 1, we get
$$2^{k+n} - (2^{2k} - 1)y - (2^k - 1)c$$
is a power of two for any $y \ge 2$ and any sufficiently large $n$. This is only possible if, for all $y \ge 2$,
$$(2^{2k} - 1)y + (2^k - 1)c = 0$$
$$\iff (2^k - 1)((2^k + 1)y + c) = 0$$
which can only hold for all $y \ge 2$ if $2^k = 1$, i.e., $f(y) = y + c$ for all $y \ge 2$. But Claim 1 for $y = 1$ and large $n$ gives
$$2^n - f(1) + 1 + c$$
is a power of two for all sufficiently large $n$, which is only possible if $f(1) = 1 + c$. Therefore $f(x) = x + c$ for all $x \in \mathbb{N}$.

We claim that all such functions are of the form $f(n) = n + c$ for some natural number $c$. The verification is trivial; now consider some function $f$ satisfying the given condition. For $n \in \mathbb{N}$, let $d(n)$ denote the number of 1's in the binary representation of $n$.

Claim 1 Given any $N \in \mathbb{N}$, there exists $m \in \mathbb{N}$ so that $f(m) > N$.

Proof. Suppose that there is $N \in \mathbb{N}$ so that $f(m) < N$ for all $m \in \mathbb{N}$. Let $k \in \mathbb{N}$ be such that $2^k > N + 1$. Note that $d(f(2^{2k} - 2^k) + 1) = d(2^{2k} - 2^k + f(1))$. The right hand side is at least $k + 1$ which the left hand side is at most $k$, a contradiction. This proves our claim. $\square$

Claim 2 The function $f$ is injective.

Proof. Suppose there are $a, b, c \in \mathbb{N}$ so that $f(a) = f(b) = c$. Then, for any natural number $k$ with $2^k > c$, we have $1 = d(f(a) + 2^k - c) = d(a + f(2^k - c))$. This implies $f(2^k - c) = 2^r - a$ for some $r \in \mathbb{N}$. Similarly $f(2^k - c) = 2^s - b$ for some $s \in \mathbb{N}$. This implies that $a - b = 2^r - 2^s$. Therefore exist unique $r$ and $s$ such that $f(2^k - c) = 2^r - a = 2^s - b$ for all natural numbers $k$ with $2^k > c$. Let $n$ be a natural number such that $f(n) - c > 2c$, which exists by the previous lemma. Let $k \in \mathbb{N}$ be such that $2^k \le f(n) - c < 2^{k+1}$ (note that this ensures $2^k > c$). Then $d(2^k + f(n) - c) = d(f(n) - c)$, but $d(2^{k+1} + f(n) - c) = d(f(n) - c) + 1$. This is a contradiction because
$$d(2^k - c + f(n)) = d(f(2^k - c) + n) = d(f(2^{k+1} - c) + n) = d(2^{k+1} - c + f(n)).$$
Thus $f$ must be injective. $\square$

Now let $a, b$ be natural numbers so that $f(a) = b$. As in the proof of the previous lemma, for any natural number $k$ with $2^k > b$ we have $f(2^k - b) = 2^r - a$ for some $r \in \mathbb{N}$.

Claim 3 There exists a natural number $K$ such that for all $k > K$, we have $f(2^k - b) = 2^k - a$.

Proof. Suppose the contrary. Let $k, r$ be distinct natural numbers so that $f(2^k - b) = 2^r - a$. Further choose distinct $l, s$ different from $k$ and $r$, so that $f(2^l - b) = 2^s - a$. Then
$$d(f(2^k - b) + 2^l - b) = d(2^r + 2^l - (a + b)) = f(2^k - b + f(2^l - b)) = d(2^s + 2^k - (a + b)). \quad (1)$$
Since $f$ is injective, we may further assume $2^{\min(k,l,r,s)} > a + b$. Note that if $m \in \mathbb{N}$ is such that $2^m > a + b$, then $d(2^{m+1} - (a+b)) = d(2^m - (a+b)) + 1$. Thus equation (1) cannot hold, proving our claim. $\square$

Let $k$ be a natural number such that $f(2^k - b) = 2^k - a$ and $2^k > 4|f(a - 1) - b|$. Note that
$$k = d(2^k - 1) = d(f(2^k - b) + a - 1) = d(2^k - b + f(a - 1)).$$
If $b > f(a-1)$ then $2^k - b + f(a-1) < 2^k$, and so we must have $f(a-1) = b-1$. On the other hand, if $f(a-1) > b$, then $< 2^k < 2^k - b + f(a-1) < 2^k + 2^{k-2}$, and hence $d(2^k - b + f(a-1)) < k-1$, a contradiction. Thus $f(a-1) = b-1$.

Letting $c = f(a) - a = b - a$, we have shown that $f(a) = a + c$ implies $f(a - 1) = a - 1 + c$, and that there are arbitrarily large integers $n$ so that $f(n) = n + c$. It is now clear by induction that $f(n) = n + c$ for all $n$. $\square$