Let $z \in \mathbb{C}$ and $n$ an integer $>3$. Define $U$ as the set of all $\lambda \in \mathbb{C}$ such that $\lambda^n = 1$.
I aim to prove the following implication: $ (\forall \lambda \in U; (\lambda -1)z + 1 \in U \cup \{0\}) \implies z = 0 \quad \text{or} \quad z = 1 $
I am uncertain if this statement is true.
The result is true for $n \ge 3$. We assume $z \ne 0$ since that works as noted and we need to prove that $z=1$. Note that for a fixed $z \ne 0$ all $(\lambda -1)z + 1$ are distinct.
Note that for any root of unity $\lambda \ne 1$ of order $n$ with argument $\theta \in (-\pi,\pi]$ we have that $\arg (\lambda-1)=\frac{\theta+\pi}{2}$. Since $n \ge 3$ there is at least one such $\lambda \ne 1$ st $(\lambda-1)z=\delta-1$ for some $\delta \in U$ (since we can take $0$ only once and there are at least two $\lambda \ne 1$ in $U$). In particular modulo $2\pi$ we have that $\arg z=\arg (\delta-1)-\arg(\lambda-1)=\frac{\pi k}{n}$ for some integral $k$
Now if $(\lambda -1)z=-1$ we get that (again modulo $2\pi$) $\arg z = \pi-\pi/2-\pi k_1/n$ and that cannot be $\pi k/n$ unless $n$ even so if $n$ odd we cannot take $0$ and $(\lambda-1)z+1 \in U$.
But then when $\lambda=\lambda_2, ...\lambda_n$ all the roots distinct from $1$ we have $(\lambda_k-1)z$ runs through all $\lambda_k-1$ also so multiplying and simplifying $\Pi_{2 \le k\le n}(\lambda_k-1)$ we get $z^{n-1}=1$ and that contradicts the fact that the argument of $z$ is $k\pi/n$ unless of course $z=1$ and the argument is zero!
Now if $n$ even and we do not take $0$ the same argument applies to give us $z=1$ Assume now that in the even case $n \ge 4$ we have a $z \ne 0,1$ solution and we show that this leads to a contradiction. Note first that $(-1-1)z+1$ must be in $U$ since otherwise $z=1/2$ but for $\lambda=e^{2\pi i/n} \ne -1$ (as $n \ge 4$) we do not have $(\lambda-1)/2+1 \in U$ since for example $|\lambda+1|/2 <1$
Since $z \ne 1$ we then have $(-1-1)z=\lambda-1$ for some $\lambda \ne -1$ so $|\lambda-1| <2$ hence $|z|<1$. But now as noted we must take $0$ so there is a root, let's renumber it $\lambda_1$ st $(\lambda_1-1)z=-1$; then if for all other roots $(\lambda-1)z \ne (\lambda_1-1)$ by counting we get that the $n-2$ other roots group both left and right so multiplying and simplifying the product of $\lambda-1$ gives us $z^{n-2}=1$ which is a contradiction with $|z|<1$.
So there must be a $\lambda_2$ st $(\lambda_2-1)z = (\lambda_1-1)$. Iterating the above, we either stop at some point and then by taking the rest of the roots which group both left and right we get $z^k=1$ for some $k \ge 1$ which again is a contradiction, or we can group the roots (that are not $1$ of course) st $(\lambda_{k+1}-1)z=\lambda_k-1, 1 \le k \le n-2$ and $(\lambda_1-1)z=-1$.
If $-1$ appears on the RHS at some point so there is $(\lambda_{k+1}-1)z=-2$, taking absolute values this is impossible since $|z|<1, |\lambda_{k+1}-1| \le 2$.
Otherwise $\lambda_{n-1}=-1$ and multiplying again all and simplifying we get $-2z^{n-1}=-1$ so $z^{n-1}=1/2$ and this gives that the argument of $z$ is $2\pi k/(n-1)$ which is a contradiction with the fact that the argument of $z$ must be $\pi k_1/n$ unless $z$ is the positive root of order $n-1$ of $1/2$ and then it is easily seen that for $\lambda=e^{2\pi i/n}$ we cannot have $(\lambda-1)z+1=\lambda z +(1-z) \in U$ since for example $|\lambda z +(1-z)|< z+1-z=1$ as $\lambda z$ and $1-z$ are not collinear. So we are finally done!