Are all algebraic integers with absolute value 1

Step 1
Let me first mention an example in Character Theory. Let G be a finite group of order n and assume ${\displaystyle \rho }$ is a representation with character ${\displaystyle \chi ={\chi }_{\rho }}$ which is defined by ${\displaystyle \chi \left(g\right)=Tr\left(\rho \left(g\right)\right)}$. Since G is a finite group then, by invoking facts from linear algebra, one can show ${\displaystyle \chi \left(g\right)\in \mathbb{Z}\left[{\zeta }_n\right]}$. For abelian groups, it is easy to see ${\displaystyle \chi \left(g\right)}$ is a root of unity, when ${\displaystyle \chi }$ is irreducible, but what about non-abelian groups? In other words let ${\displaystyle \left|\chi \left(g\right)\right|=1}$, what can we say about ${\displaystyle \chi \left(g\right)}$?
This relates to your question. Let assume ${\displaystyle \frac{K}{\mathbb{Q}}}$ be an abelian Galois extension inside ${\displaystyle \mathbb{C}}$, and take an algebraic integer ${\displaystyle \alpha \in {\mathcal{O}}_K}$ such that ${\displaystyle \left|\alpha \right|=1}$, then for any ${\displaystyle \sigma \in Gal\left(\frac{K}{\mathbb{Q}}\right)}$ we have
${\displaystyle {\left|\sigma \left(\alpha \right)\right|}^2=\sigma \left(\alpha \right)\overline{\sigma \left(\alpha \right)}}$
Since ${\displaystyle \frac{K}{\mathbb{Q}}}$ is abelian then
${\displaystyle \overline{\sigma \left(\alpha \right)}=\sigma \left(\bar{\alpha }\right)}$ so
${\displaystyle {\left|\sigma \left(\alpha \right)\right|}^2=\sigma \left(\left|\alpha \right|\right)=1}$
Then norm of all its conjugate is one so it must be a root of unity. This answer to the question was posed, therefore if ${\displaystyle \left|\chi \left(g\right)\right|=1}$ then ${\displaystyle \chi \left(g\right)}$ is root of unity.

Step 1
Let x be an algebraic number with absolute value 1. Then x and its complex conjugate ${\displaystyle \bar{x}=\frac{1}{x}}$ have the same minimal polynomial.
Writing f(T) for the minimal polynomial of x over ${\displaystyle \mathbb{Q}}$, with degree n, the polynomials ${\displaystyle T^{n}f\left(\frac{1}{T}\right)}$ and ${\displaystyle f\left(T\right)}$ are irreducible over ${\displaystyle \mathbb{Q}}$ with root ${\displaystyle \bar{x}}$, so the polynomials are equal up to a scaling factor:
${\displaystyle T^{n}f\left(\frac{1}{T}\right)=cf\left(T\right)}$.
Setting ${\displaystyle T=1,f\left(1\right)=cf\left(1\right)}$
Assuming x is not rational (i.e., x is not 1 or -1), f has degree greater than 1, so f(1) is nonzero and thus ${\displaystyle c=1}$. Therefore
${\displaystyle T^{n}f\left(\frac{1}{T}\right)=f\left(T\right)}$,
so f(T) has symmetric coefficients. In particular, its constant term is 1. Moreover, the roots of f(T) come in reciprocal pairs (since 1 and -1 are not roots), so n is even.
Partial conclusion: an algebraic number other than 1 or -1 which has absolute value 1 has even degree over ${\displaystyle \mathbb{Q}}$ and its minimal polynomial has constant term 1. In particular, if x is an algebraic integer then it must be a unit.
There are no examples of algebraic integers with degree 2 and absolute value 1 that are not roots of unity, since a real quadratic field has no elements on the unit circle besides 1 and -1 and the units in an imaginary quadratic field are all roots of unity (and actually are only 1 and -1 except for ${\displaystyle \mathbb{Q}\left(i\right)}$ and ${\displaystyle \mathbb{Q}\left(\omega \right)}$). Thus the smallest degree x could have over ${\displaystyle \mathbb{q}}$ is 4 and there are examples with degree 4: the polynomial
${\displaystyle x^4-2x^3-2x+1}$
has two roots on the unit circle and two real roots (one between 0 and 1 and the other greater than 1).