Let us consider the polynomial ring $\Bbb C[x\_1,...,x\_s]$ and $\alpha(x\_i)= x\_i + \mu\_i$ where $\mu\_i \in \Bbb C$ are not all zero. Then $\alpha \in \mathrm{Aut}(\Bbb C[x\_1,...,x\_s])$. 



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> What are the fixed points of $\alpha^n-\mu\_j$ for a fixed $j$, i.e. what are the $a\_j \in \Bbb C[x\_1,...,x\_s]$ s.t. $\alpha^n(a\_j)-\mu\_j=a\_j$? 
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You can say some condition on $\mu\_i$'s too so that we can get a fixed point.


Next do the same for $\Bbb C[x\_1^{\pm 1},...,x\_s^{\pm1}]$


It is evident that any constant polynomial will not be fixed.


I have tried but can't find any particular result except that "If $\mu\_i$'s will be algebraically independent (i.e. no polynomial relation in between them) then no fixed point would be there." But this is almost trivial as soon as you write explicit expression.



What are the fixed points of $\alpha^n-\mu_j$ for a fixed $j$?
Let $L$ be the line generated by $\underline{\mu} = (\mu\_1,\dots,\mu\_s)$ in $V = \mathbb{C}^s$, and consider a projection $p : V \rightarrow V$ onto $L$, with kernel $H$. Any polynomial function of the form
$$
a\_j(\underline{x}) = b(\underline{x} - p(\underline{x})) + \ell\_j( p (\underline{x})),
$$
where $b$ is a polynomial function on $H$, and where $\ell\_j$ is the linear form on $L$ such that $\ell(n \underline{\mu}) = \mu\_j$, is a solution to your equation.


Conversely, any fixed point $a\_j$ has this form. Indeed, if $\underline{y}$ is an element of $H$, then an easy induction on $t$ yields the formula
$$
a\_j(\underline{y} + tn \underline{\mu}) = a\_j(\underline{y}) +t\mu\_j
$$
for any integer $t$, so that the polynomial $a\_j(\underline{y} + T n \underline{\mu}) - a\_j(\underline{y}) - \ell\_j(T n \underline{\mu})$ has infinitely many roots, and is thus zero. This yields the result with $b = {a\_j}\_{|H}$.


