If $0$ is in $\mathbb{N}$ with the convention $\varphi(0)=0$, then there is a unique solution $f(n)=0$ for every $n\in\mathbb{N}$. From now on, I assume that $\mathbb{N}$ is defined to be $\{1,2,3,\ldots\}$. I claim that there does not exist such a function $f$.
Suppose on the contrary that $f$ exists. Let $g:=\varphi\circ f$. Then, $g$ satisfies Cauchy's functional equation. That is, there exists $c\in\mathbb{N}$ such that $g(n)=cn$ for all $n\in\mathbb{N}$.
Let $d_1,d_2,\ldots,d_k$ be all natural numbers that divide $c$. Pick pairwise distinct primes $p_1,p_2,\ldots,p_{2k}$ that do not divide $c$. Consider the system of congruences
$$d_ix\equiv-1\pmod{p_{2i-1}p_{2i}}$$
for $i=1,2,\ldots,k$. This congruence has a unique solution
$$x\equiv x_0\pmod{P}\,,$$
where $P:=p_1p_2\cdots p_{2k}$ and $x_0\in\mathbb{Z}_{>0}$. By Dirichtlet's Theorem, there exists a prime natural number $p>c$ such that $p\equiv x_0\pmod{P}$, noting that $\gcd(x_0,P)=1$. It follows that the equation
$$\varphi(N)=cp$$
has no solution $N\in\mathbb{N}$. Thus, $\varphi\big(f(p)\big)=g(p)=cp$ is impossible.
We first consider the case $c \leq 1/4$; we shall show in this case
$f$ must be constant. The relation
$$
f(x) = f(x^2 + c) = f((-x)^2 + c) = f(-x)
$$
proves that $f$ is an even function. Let $r_1 \leq r_2$ be the roots of
$x^2 + c - x$, both of which are real. If $x > r_{2}$, define $x_{0} =
x$ and $x_{n+1} = \sqrt{x_{n} - c}$ for each positive integer $x$. By
induction on $n$, $r_{2} < x_{n+1} < x_{n}$ for all $n$, so the
sequence $\{x_{n}\}$ tends to a limit $L$ which is a root of $x^{2} +
c = x$ not less than $r_{2}$. Of course this means $L = r_{2}$.
Since $f(x) = f(x_{n})$ for all $n$ and $x_{n} \to r_{2}$, we
conclude $f(x) = f(r_{2})$, so $f$ is constant on $x \geq r_{2}$.
If $r_{1} < x < r_{2}$ and $x_{n}$ is defined as before, then by
induction, $x_{n} < x_{n+1} < r_{2}$. Note that the
sequence can be defined because $r_{1} > c$; the latter follows by
noting that the polynomial $x^{2} - x + c$ is positive at $x = c$ and
has its minimum at $1/2 > c$, so both roots are greater than $c$. In
any case, we deduce that $f(x)$ is also constant on $r_{1} \leq x \leq
r_{2}$.
Finally, suppose $x < r_{1}$. Now define $x_{0} = x, x_{n+1} =
x_{n}^{2} + c$. Given that $x_{n} < r_{1}$, we have $x_{n+1} >
x_{n}$. Thus if we had $x_{n} < r_{1}$ for all $n$, by the same argument as
in the first case we deduce $x_{n} \to r_{1}$ and so $f(x) =
f(r_{1})$. Actually, this doesn't happen; eventually we have $x_{n} >
r_{1}$, in which case $f(x) = f(x_{n}) = f(r_{1})$ by what we have
already shown. We conclude that $f$ is a constant function. (Thanks
to Marshall Buck for catching an inaccuracy in a previous version of
this solution.)
Now suppose $c > 1/4$. Then the sequence $x_n$ defined by $x_0 = 0$
and $x_{n+1} = x_n^2 + c$ is strictly increasing and has no limit
point. Thus if we define $f$ on $[x_0, x_1]$ as any continuous
function with equal values on the endpoints, and extend the definition
from $[x_n, x_{n+1}]$ to $[x_{n+1}, x_{n+2}]$ by the relation $f(x) =
f(x^2 + c)$, and extend the definition further to $x < 0$ by the
relation $f(x) = f(-x)$, the resulting function has the desired
property. Moreover, any function with that property clearly has this form.
Best Answer
Claim: The continuous functions satisfying the equation $f(x) = f(x^af(x))$ are exactly the set of functions of the following form for some constants $c_1,c_2$ such that $0\le c_1\le c_2\le \infty$:$$ f(x) =\begin{cases}\frac1{c_1^{a-1}} & x \le c_1 \\ \frac1{x^{a-1}} &x\in(c_1,c_2)\\\frac1{c_2^{a-1}} & x \ge c_2 \end{cases} $$
This is clearly continuous on $(0,\infty)$ and it satisfies the functional equation trivially for $x\in(c_1,c_2)$. For $x\le c_1$, we note that $c_1\ge \frac{x^a}{c_1^{a-1}}=x^af(x)$ so we have $\frac1{c_1^{a-1}}=f(x)=f(x^af(x))$. Similarly this $f$ works for $x\ge c_2$. (Note: $f$ is constant if $c_1=c_2$, and $f(x) = \frac1{x^{a-1}}$ if $c_1=0$ and $c_2=\infty$.)
To prove the claim, we'll make use of what I'm going to call the reflection princple: If $f$ satisfies the functional equation, then so does $\bar f(x) = \frac1{f(1/x)}$.
Define $h(x) =x^a f(x)$. Then we have $f(x) = f(h(x))$, and $h$ is a continuous function from $(0,\infty)$ to $(0,\infty)$. We will prove the claim as a series of lemmas about $f$ and $h$.
Lemma 1: $h(x)$ is strictly increasing.
Proof: First, we show $h$ is injective. Suppose $h(x)=h(y)$. Then \begin{eqnarray} &&h(x)=h(y)\\ &\implies& f(h(x))=f(h(y))\\ &\implies& f(x) = f(y)\\ &\implies& \frac{h(x)}{x^a} = \frac{h(y)}{y^a}\\ &\implies& x=y \end{eqnarray} Thus $h$ is either strictly increasing or strictly decreasing. If it is decreasing, we may derive a contradiction, because in that case $f(x)=h(x)/x^a$ would also be strictly decreasing (as a product of two decreasing positive functions), and thus in particular $f$ is invertible and therefore $f(x) =\frac1{x^{a-1}}$, which implies $h(x) = x$, which is not decreasing.
Lemma 2: The set of points such that $h(x)=x$ is a closed interval in $(0,\infty)$. [Note: In the subspace topology, $(0,\infty)$ and intervals of the form $(0,b]$ are closed though they are not closed in $\mathbb R$.]
Proof: We use contradiction to prove the slightly weaker claim that if $A<B$ are real numbers with $h(A) = A$ and $h(B)=B$, then there exists $C\in(A,B)$ such that $h(C)=C$.
Suppose $h$ has no fixed points in $(A,B)$. By the reflection principle, we can suppose without loss of generality that $h(x)>x$ for all $x\in(A,B)$. By injectivity and continuity of $h$, we can see that it is a bijection from $(A,B)$ to itself. Furthermore, it is evident that, for each $x\in(A,B)$, the sequence $\{x,h(x),h(h(x)),\dots\}$ converges to $B$. Therefore, for all $x\in(A,B)$, we have $$ f(x) = f(h(x)) = f(h(h(x))) = \lim\limits_{n\rightarrow\infty} f(h^n(x)) = f(B) = B $$ But this makes $f$ discontinuous at $A$, a contradiction. Therefore $f$ has a fixed point in $(A,B)$.
This implies that, if $A\le B$ are fixed points of $h$, the fixed points of $h$ are dense in $[A,B]$. By continuity of $h$, the set of fixed points is a closed set, and therefore is cannot be anything other than a closed subinterval of $(0,\infty)$.
Lemma 3: $h(x)$ has a fixed point.
Proof: We again prove by contradiction. Suppose $h(x)\ne x$ for all $x$. By the reflection principle, we may suppose without loss of generality that $h(x)>x$ for all $x$. This implies for any $x$ $$ \lim\limits_{n\rightarrow\infty} h^n(x) = \infty $$ Fix $x_0$ and define $I_n = [h^n(x_0),h^{n+1}(x_0)]$. Let $M=\max_{x\in I} f(x)$. We note that $h(I_{n-1}) = I_n$, thus for each $n$, $f(x)\le M$ for all $x\in I_n$. Because $[x_0,\infty)=\bigcup_{n\ge0} I_n$, we conclude $f(x)\le M$ for all $x\in [x_0,\infty)$. However, this contradicts $h(x)>x$ for all $x$: For $n$ sufficiently large, we have $h^n(M^{-\frac1{a-1}} )\ge x_0$, and thus $$ M^{-\frac1{a-1}} < h(M^{-\frac1{a-1}}) = M^{-\frac a{a-1}}f(M^{-\frac1{a-1}}) = M^{-\frac a{a-1}}f(h^n(M^{-\frac1{a-1}})) \le M^{-\frac a{a-1}} M=M^{-\frac1{a-1}}, $$ a contradiction.
Lemma 4: Suppose $h(B)=B$ and $h(x)\ne x$ for all $x>B$. Then $f(x)$ is constant on $[B,\infty)$.
Proof: By continuity, we have either $h(x)>x$ or $h(x)<x$ for all $x>B$. In the former case, we observe that $h$ is a bijection on $(B,\infty)$, and thus we have a $h^{-1}$ is well defined with $h^{-1}(x)<x$ for all $x>B$. We also note that $f(h^{-1}(x)) = f(x)$ for all $x$.
Pick $\epsilon>0$. Then there exists $B'>B$ such that, for all $x\in(B,B')$, $|f(x) - f(B)|<\epsilon$. If we pick $x>B$ arbitrary, we observe that either $h^n(x)$ decreases to $B$, or $h^{-n}(x)$ decreases to $B$ (depending on whether $h(x)<x$ or $h(x)>x$ for $x>B$). Thus, we have for some $z\in\mathbb Z$, $h^z(x) \in(B,B')$. Therefore $$ |f(x)-f(B)| = |f(h^z(x))-f(B)|<\epsilon $$ Since $x>B$ was arbitrary, we conclude $|f(x)-f(B)| <\epsilon$ for all $x>B$. Letting $\epsilon$ go to $0$, we conclude $f(x)=f(B)$ for any $x>B$.
Proof of claim: Let $f(x)$ be a continuous function satisfying $f(x)=f(x^a f(x))$ for all $x\in(0,\infty)$. It is straightforward from here to show that $f$ has the desired form. We note that $f(x)=\frac1{x^{a-1}}$ is equivalent to $x$ being a fixed point of $h$. By Lemmas 2 and 3, there exists a closed interval $I$ such that $f(x) = \frac{1}{x^{a-1}}$ for all $x\in I$, and $f(x)\ne \frac{1}{x^{a-1}}$ for all $x\notin I$. By Lemma 4, if $f$ is constant to the right of $I$. Suppose the left endpoint of $I$ is $A>0$. By the reflection principle, we have that $\bar{f}$ is equal to $\frac1{x^{a-1}}$ in the interval $1/I$. Thus, by Lemma 4, $\bar{f}$ is constant for $x>\frac1A$, which implies that $f$ is constant for $x\in(0,A]$. Using continuity of $f$, we can see what the constant values are to the right and left of $I$ and therefore $f$ must have the claimed form for some $c_1,c_2$. Q.E.D.