Consider all bijective mappings from $G \to G$ ($G$ is a finite cyclic group) in which generators are mapped only to generators. Are all these mappings isomorphisms?
I know that an isomorphism between $2$ cyclic groups takes a generator to generator. I want to know whether the reverse is true as well if given that the mapping is a bijection.
On
No. Consider $G = \mathbb{Z} / 5 \mathbb{Z}$. Every element except $0$ is a generator. There are only $4$ automorphisms of $G$ since a group homomorphism $G \to G$ is entirely determined by where it sends $1$, since $1$ generates $G$, and you can’t send $1$ to $0$. In other words, the only automorphisms of $G$ are the ones that multiply by $1, 2, 3,$ or $4$.
But there are $4! = 24$ different permutations of the generators $\{ 1, 2, 3, 4 \}$. So there are $24-4=20$ counterexamples.
EDIT: In general, if $p$ is prime, then $\mathbb{Z} / p \mathbb{Z}$ has $(p - 1)$ generators, has $(p - 1)$ automorphisms, and has $(p - 1)!$ permutations of its generators.
On
To answer all questions about it. Let $G$ be a cyclic group of order $n$. Then the number of automorphisms of group $G$ equals $\phi(n)$, where $\phi(n)$ is an Euler function. The number of bijections of $G$ to itself which generators of $G$ are mapped only to generators of $G$ is equal to $$ \phi(n)!(n-\phi(n))! $$ The inequality $$ \phi(n)!(n-\phi(n))!>\phi(n) $$ is satisfied if and only if $n\geq4$.
No. Consider the cyclic group $G=\mathbb Z/8\mathbb Z$ under addition. Generators are $1, 3, 5, $ and $7$, but if a bijection maps $2\mapsto4$, it's not an isomorphism, because it doesn't preserve the order of all of the elements.