We show that the only such continous function is $f(x) = x$ for all $x \in \mathbb{R}$. Observe that this obviously satisfies the problem's constraints. Let us now show this is the only valid function. Write $N = k!$ where $k = \lfloor \varphi^{2023} \rfloor$. We first claim that $f^N(x) = x$ for all $x \in \mathbb{R}$ except possibly, for an interval of length at most $2 \varphi$. First, we show that $f$ has a fixed point. If this were not the case, then we would either have $f(x) > x$ for all $x \in \mathbb{R}$ or $f(x) < x$ for all $x \in \mathbb{R}$ because $f$ is continous. Take any $x, y$ with $|x-y| > \varphi$. We have for some $n \in \mathbb{N}$ the equality $f^n(x)-x = -(f^n(y)-y)$. It follows by continuity that $f^n(r) = r$ for some $r \in [x, y]$. But because $f(x) > x$ for all $x \in \mathbb{R}$ or $f(x) < x$ for all $x \in \mathbb{R}$ this is impossible. Thus $f$ has a fixed point, say $a$. Consider the interval $[a-\varphi, a+\varphi]$. Any real $r$ outside this interval satisfies $|r-a| > \varphi$ and thus has some $n$ corresponding to it such that $f^n(r)+f^n(a) = r+a$ or that $f^n(r) = r$ implying that $f^N(r) = r$. Hence, our claim is justified. Now we observe that if $f(x) = f(y)$ for some $x, y$ not in the interval $I \overset{\text{def}}{=} [a-\varphi, a+\varphi]$ we have $x = y$ because $f^N(x) = x$ and $f^N(y) = y$. Because $f$ is injective and continous over $(a+\varphi, \infty)$ it follows that $f$ is monotonous over it. Similarily, $f$ is monotonous over $(-\infty, a-\varphi)$. Since $\lim_{k \mapsto \infty} f(k) = \infty$, it follows that $f$ is increasing over $(a+\varphi, \infty)$. Observe that this implies $f(x)$ is also increasing over $(-\infty, a-\varphi)$ as otherwise, we would lose injectivity. If $f(x) > x$ for some $x \in (a+ \varphi, \infty)$ then the sequence $\{f^i(x)\}_{i \geq 1}$ is unbounded, contradicting $f^N(x) = x$. So $f(x) \leq x$ over this interval. Let $M$ be an upper bound over all $f(x)$ with $x < a+\varphi$. Pick $x > M$ and consider $\{f^i(x)\}_{i \geq 1}$. Observe that each term of this sequence is always less than $x$, unless $f(x) = x$. Since $f^N(x) = x$, the latter must necessarily hold. It follows that $f(x) = x$ for all $x > M$. Pick some $x$ large enough and observe that we also now have $f^N(x) = x$ for all $x \in I$ (selecting $x > \max(M, a+2\varphi)$ suffices). In particular, this extends monotonicity throughout $\mathbb{R}$. Now pick any $x \in \mathbb{R}$. If $f(x) > x$ then $f^i(x)$ grows unbounded, contradicting $f^N(x) = x$. If $f(x) < x$ then again $f^i(x)$ decreases unbounded contradicting $f^N(x) = x$. Thus $f(x) = x$ at each point and we are finally done.