Suppose $\varphi:\mathbb{N}\rightarrow\mathbb{N}$ is a finite-to-one map. We can then define a function $\varphi\_1:a\mapsto |\varphi^{-1}(a)|$. If this function is finite-to-one, we can similarly define $\varphi\_2=(\varphi\_1)\_1: a \mapsto |\varphi\_1^{-1}(a)|$ and so on. 


For example, the identity function on $\mathbb{N}$ would have its first induced function be the constant function 1, and thus would not have a second induced function. 


Question: Is there a function $\varphi$ such that $\varphi\_n$ exists for all $n$? If so, is there any obvious classification (i.e. necessary and sufficient conditions)?


EDIT: If the answer is yes, then for (possibly) an added challenge, can we make all the $\varphi\_n$ surjective?



"Stably" finite-fiber maps of the integers
If $\mathbb N$ starts with $0$ then you can't have both $\phi\_n$ and $\phi\_{n+1}$ surjective. That would mean there exists $a$ such that $\phi\_{n+1}(a)=0$, so that $a$ is not in the image of $\phi\_n$.


But if we use functions from positive integers to positive integers then we can even make a solution for which $\phi\_n$ is independent of $n$. Define $\phi(1)=1$, $\phi(2)=\phi(3)=2$, $\phi(4)=\phi(5)=3$, $\phi(6)=\phi(7)=\phi(8)=4$, and so on: when you come to picking which numbers go to $k$, just take the next $j$ in a row where $j$ is $\phi(k)$ and has already been defined.


