= Solution
A <group homomorphism> preserves multiplication, $\phi(ab)=\phi(a)\phi(b)$. A <group isomorphism> is a bijective homomorphism. An isomorphism from a group to itself is a <group automorphism>.
The <automorphism group> contains the identity map. Composition is associative, and if $\phi,\psi$ are automorphisms, $(\phi\circ\psi)(ab)=\phi(\psi(a)\psi(b))=(\phi\circ\psi)(a)(\phi\circ\psi)(b)$; composition is also bijective. For $u=\phi(a)$ and $v=\phi(b)$, $\phi^{-1}(uv)=\phi^{-1}(\phi(ab))=ab=\phi^{-1}(u)\phi^{-1}(v)$, proving the inverse bijection is a homomorphism. These facts establish all group axioms.
A finite <cyclic group> is $G=\langle g\rangle=\{e,g,\ldots,g^{n-1}\}$ with $g$ of order $n$. The map $g^k\mapsto e^{2\pi ik/n}$ is well defined modulo $n$, bijective onto the <roots of unity>, and preserves multiplication. It identifies $G$ with the multiplicative group of $n$th roots of unity.
For prime $n$, <Lagrange's theorem> leaves only subgroup orders one and $n$: \b[there is no nontrivial proper subgroup]. The trivial subgroup $\{e\}$ is nevertheless proper when $n>1$, so that qualification is necessary in the printed claim. The characterization of <groups with no proper nontrivial subgroups> makes the distinction explicit.
For composite $n$, choose a prime divisor $p$ and take $\boxed{H_0=\{e\},\quad H_1=\langle g^p\rangle}$, of orders $1$ and $n/p$. They are distinct proper subgroups, with $H_1$ nontrivial. If the question intended two nontrivial proper subgroups, that demand is not always possible: a cyclic group of order $p^2$ has only the one of order $p$. The <subgroups and quotients of a cyclic group> are controlled by its divisors. The order-one group is a separate trivial case, with no proper subgroups.
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