Problem: Find all nonconstant polynomials $P(z)$ with complex coefficients
for which all complex roots of the polynomials
$P(z)$ and $P(z)-1$ have absolute value $1$.

Solution: The answer is $P(x)$ should be a polynomial
of the form $P(x) = \lambda x^n - \mu$
where $|\lambda| = |\mu|$ and $\opname{Re} \mu = -\half$.
One may check these all work;
let's prove they are the only solutions.

\paragraph{First approach (Evan Chen).}
We introduce the following notations:
\begin{align*}
  P(x) &= c_n x^n + c_{n-1}x^{n-1} + \dots + c_1 x + c_0 \\
    &= c_n (x+\alpha_1) \dots (x+\alpha_n) \\
  P(x)-1 &= c_n (x+\beta_1) \dots (x+\beta_n)
\end{align*}
By taking conjugates,
\begin{align*}
  (x + \alpha_1) \dotsm (x + \alpha_n) &= (x + \beta_1) \dotsm (x + \beta_n) + c_n^{-1} \\
  \implies \left(x + \frac{1}{\alpha_1}\right) \dotsm \left(x + \frac{1}{\alpha_n}\right)
  & = \left(x + \frac{1}{\beta_1}\right) \dotsm \left(x + \frac{1}{\beta_n}\right)
    + (\ol{c_n})^{-1} \qquad (\spadesuit)
\end{align*}
The equation $(\spadesuit)$ is the main player:
\begin{claim*}
  We have $c_k = 0$ for all $k = 1, \dots, n-1$.
\end{claim*}
\begin{proof}
  By comparing coefficients of $x^k$ in $(\spadesuit)$
  we obtain
  \[ \frac{c_{n-k}}{\prod_i \alpha_i} = \frac{c_{n-k}}{\prod_i \beta_i} \]
  but $\prod_i \alpha_i - \prod_i \beta_i = \frac{1}{c_n} \neq 0$.
  Hence $c_k = 0$.
\end{proof}

It follows that $P(x)$
must be of the form $P(x) = \lambda x^n - \mu$,
so that $P(x) = \lambda x^n - (\mu + 1)$.
This requires $|\mu| = |\mu+1| = |\lambda|$
which is equivalent to the stated part.

\paragraph{Second approach (from the author).}
We let $A = P$ and $B = P-1$ to make the notation more symmetric.
We will as before show that $A$ and $B$ have all coefficients equal to zero
other than the leading and constant coefficient; the finish is the same.

First, we rule out double roots.
\begin{claim*}
  Neither $A$ nor $B$ have double roots.
\end{claim*}
\begin{proof}
  Suppose that $b$ is a double root of $B$.
  By differentiating, we obtain $A' = B'$,
  so $A'(b) = 0$.
  However, by Gauss-Lucas,
  this forces $A(b) = 0$, contradiction.
\end{proof}
Let $\omega = e^{2\pi i/n}$,
let $a_1, \dots, a_n$ be the roots of $A$,
and let $b_1, \dots, b_n$ be the roots of $B$.
For each $k$,
let $A_k$ and $B_k$ be the points in the complex plane
corresponding to $a_k$ and $b_k$.
\begin{claim*}[Main claim]
  For any $i$ and $j$,
  $\tfrac{a_i}{a_j}$ is a power of $\omega$.
\end{claim*}
\begin{proof}
  Note that
  \[ \frac{a_i - b_1}{a_j - b_1} \dotsm \frac{a_i - b_n}{a_j - b_n}
    = \frac{B(a_i)}{B(a_j)} = \frac{A(a_i) - 1}{A(a_j) - 1} = \frac{0-1}{0-1} = 1. \]
  Since the points $A_i$, $A_j$, $B_k$ all lie on the unit circle,
  interpreting the left-hand side geometrically gives
  \[ \dang A_iB_1A_j + \dots + \dang A_iB_nA_j = 0
    \implies n \widehat{A_iA_j} = 0, \]
  where angles are directed modulo $180\dg$
  and arcs are directed modulo $360\dg$.
  This implies that $\tfrac{a_i}{a_j}$ is a power of $\omega$.
\end{proof}
Now the finish is easy:
since $a_1$, \dots, $a_n$ are all different,
they must be $a_1 \omega^0$, \dots, $a_1 \omega^{n-1}$ in some order;
this shows that $A$ is a multiple of $x^n-a_1^n$, as needed.