= Solution
Put $U=\phi(A(t))$. Applying the <group homomorphism> $\phi$ to the identity just proved shows that $U$ is conjugate to $U^{m^2}$ for every positive integer $m$. Thus the multiset of its <eigenvalues> is unchanged by taking their $m^2$th powers.
Take $m=2$. The fourth-power operation permutes this finite multiset. For each <eigenvalue> $z$, iterating its permutation cycle gives $z^{4^r}=z$ for some positive integer $r$. Since $U$ is unitary, $z\ne0$, so $z^{4^r-1}=1$: every <eigenvalue> is a <root of unity>. Choose a positive integer $d$ divisible by all their orders. Every <eigenvalue> of $U^{d^2}$ is then one. Because its <eigenvalue> multiset is also the original one, we obtain
$$
\boxed{\text{every eigenvalue of }\phi(A(t))\text{ equals }1.}
$$
This is <square-power rigidity of a finite nonzero spectrum>. Finiteness of the dimension is essential to the permutation-cycle argument.
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