Fixed points of recursive functions with finite range Let $\phi$ be a programming system satisfying the UTM Theorem (i.e., $\phi$ is a $2$-ary partial recursive function such that the list $\phi_0,\phi_1,\ldots$ includes all $1$-ary partial recursive functions, where $\phi_i=\lambda x.\phi(i,x)$ for all $i\in\mathbb{N}$). Suppose that for all recursive functions $f$ with finite range, there exits an $n\in\mathbb{N}$ such that $\phi_n=\phi_{f(n)}$. If $g$ is any recursive function (with finite or infinite range), is there an $n\in\mathbb{N}$ such that $\phi_n=\phi_{g(n)}$.
 A: Expanding on my comment above.  There is a Friedberg listing of the partial recursive functions, i.e. a 2-ary p.r. function $\phi$ such that $\phi_i = \lambda x.\phi(i,x)$ lists every p.r. function without repetition.  We may assume that $\phi_0$ is the empty function.
Now we'll define a new list $\psi$ with $\psi_{2i} = \phi_i$ for all $i$.  We also promise that for each $2i+1$, there will be a $j < 2i+1$ with $\psi_{2i+1} = \psi_{j}$.  By induction, we may assume that $j$ is even.  Now let $g$ be any total computable function such that $g(n)$ is always an even number strictly greater than $n$.  Then $g$ has no fixed points in this system.  (Because a fixed point of $g$ would give us $j < i$ with $\phi_j = \phi_i$.)
It remains only to define $\psi_{2i+1}$.  If $i = \langle e, k\rangle$, where this is the standard pairing function, then we let $\psi_{2i+1}$ be the empty function until $\phi_e(2i+1)\!\downarrow = j < 2i+1$.  Once this occurs, we make $\psi_{2i+1}$ copy $\psi_j$.  Then $2i+1$ is a fixed point of $\phi_e$.
If $f = \phi_e$ is total and has a bounded range, then for a sufficiently large $k$, $f(2\langle e,k\rangle+1) < 2\langle e, k\rangle+1$, so $f$ will have a fixed point.
