Assume that $G$ is a group such that a subset of $G$ like $S$ can generate $G$.
Prove that:
If $\sigma$ is an automorphism in $G$, Then $\sigma(S)$ generates $G$, too.
Note: If $S$ generates $G$, it means that The cyclic group generated by the elements of $S$ is equal to $G$. So, How does an isomorphism from $G$ to itself (an automorphism called $\sigma$), become useful here? Can we say that $\sigma(S) = S$ ? (Because $\sigma$ is bijective?)
An automorphism is an isomorphism of a group between itself. Note that $\sigma$ is then a homomorphism. Since $\sigma$ is 1-to-1 in the whole group $G$, it is 1-to-1 in $S$. Then
$$\sigma : S \to \sigma(S)$$
Becomes a 1-to-1 and onto homomorphism. A group under a homomorphism is a group too so clearly we have $\sigma(S) \le G$.
Let the set $A \subset S$ generate $S$. Note that $A$ could be either a proper or improper subset of $S$. We are given that $A$ generates G. Let $A=(x_1,x_2,...,x_n)$ for any element $k \epsilon S$
$$\sigma(k)=\sigma((x_1n_1)(x_2n_2)...(x_nn_n))=\sigma(x_1)\sigma(n_1)...\sigma(x_n)\sigma(n_n)$$
Since $\sigma$ is an homomorphism.Since $\sigma$ is ONTO, clearly we can see from the equation above that for $\sigma$ to generate all elements of $\sigma(S)$ It's image of the generator should be a generator.
then $\sigma(A)=A$
The result follows.