Do the solutions to functional equation $f(f(z))=f(z-c)+c$ necessarily have a fixed point? Let a function $f(z)$ defined for all $z\in\mathbb{Z}$ (without any further restrictions) satisfy the equation:
$$
f(f(z))=f(z-c)+c,\tag{1}
$$
where $c\ne0$ is an integer constant. Is the following statement valid: 
There exists $z_0$ such that:
$$
f(z_0)=z_0.\tag{*}
$$
Provided that such a point exists it can be shown that the equation (*) is satisfied in fact for all points:
$$
z_k=z_0+kc,\quad k\in\mathbb{Z}.\tag{2}
$$
Initially the conjecture was formulated for complex domain, but appears to be false. By similar argument it should fail also in real domain, though the counterexample shall be obviously a bit trickier. However can the conjecture still be valid for integer functions?
 A: No.  Let me first consider a structure that may seem unrelated but whose relevance will be made clearer later.  Let $G$ be the group of affine automorphisms of $\mathbb{R}$: that is, maps $\mathbb{R}\to\mathbb{R}$ of the form $x\mapsto ax+b$ for $a\neq 0$.  Let $f\in G$ be the map $x\mapsto x+1$ and let $c\in G$ be the map $x\mapsto 2x$.  Note that these group elements satisfy the relation $$f^2=cfc^{-1}$$ (the choice of variables $f$ and $c$ is not a coincidence, and hopefully this is starting to look familiar!).
Now let $H$ be the subgroup of $G$ generated by $c$ and let $H$ act on $G$ by left translation.  As an $H$-set, $G$ is an disjoint union of $2^{\aleph_0}$ copies of $H$ (namely, all the cosets of $H$).
Now let $H$ also act on $\mathbb{C}$, with $c$ acting as the map $z\mapsto z+c$ (this is a well-defined action since $c$ has infinite order).  Once again, as an $H$-set, $\mathbb{C}$ is just a disjoint union of $2^{\aleph_0}$ copies of $H$.  So there is an isomorphism of $H$-sets $F:G\to \mathbb{C}$.
Now we define our map $f:\mathbb{C}\to\mathbb{C}$ by just transferring multiplication by the element $f\in G$ along the isomorphism $F$.  That is, we define $$f(z)=F(f\cdot F^{-1}(z)).$$ The relation $f^2=cfc^{-1}$ on $G$ then exactly turns into the functional equation $$f(f(z))=f(z-c)+c.$$  Moreover, there does not exist any $z\in\mathbb{C}$ such that $f(z)=z$, since that would mean that $f\cdot F^{-1}(z)=F^{-1}(z)$ which is impossible because $f$ is not the identity element of $G$.

The larger insight here is that your functional equation involves no structure of the set $\mathbb{C}$ other than the map $z\mapsto z+c$.  That is, it depends only on the structure of $\mathbb{C}$ as an $H$-set.  That structure is quite simple: it's just a huge disjoint union of copies of $H$, and the group $H$ is isomorphic to $\mathbb{Z}$.  So you can more easily construct solutions to the functional equation by forgetting all the irrelevant structure of $\mathbb{C}$ and just thinking of it as a giant disjoint union of copies of $\mathbb{Z}$, with the map $z\mapsto z+c$ corresponding to adding $1$ on each copy of $\mathbb{Z}$.
The use of the group $G$ above is a particularly elegant way to harness this to construct a counterexample to your question, but you could also construct one by just messing around with a big disjoint union of copies of $\mathbb{Z}$ directly.  For instance, you could start by picking one copy of $\mathbb{Z}$, and say that $f$ should map one element of it to a different copy of $\mathbb{Z}$, and then see what further properties of $f$ are forced by the relation $f^2=cfc^{-1}$.  With a bit of work you can work out a countable collection of copies of $\mathbb{Z}$ which $f$ can map to each other while satisfying the functional equation*.  Partitioning the $2^{\aleph_0}$ copies of $\mathbb{Z}$ into countable pieces, you can then use this recipe on each piece to get the desired $f$.
*This essentially amounts to constructing the free monoid with elements $f$ and an invertible element $c$ satisfying $f^2=cfc^{-1}$, and verifying that the subgroup generated by $c$ acts freely on it and $f$ has no fixed points.  The trick of using the group $G$ as I did is that it provides a ready-made monoid satisfying these relations with the subgroup generated by $c$ acting freely and $f$ having no fixed points, so you don't have to construct one combinatorially.
