I'm kinda stuck with this task.
Let $G$ be a group and $ S $ the set of all isomorphisms $ f: G \to G$. I first want to show that $ (S, \circ) $ is also a group.
I believe I've shown that all the properties of a group is fulfilled with $ (S, \circ) $:
i) Assume that $ x \in $ and $ f_1,f_2 \in S$. Then $f_2(x) = y \in G$, since $ f_1 $ is an ismorphism. $ f_2(y) = z \in G $ since $ f_2 $ is an ismorphism. Then $ f_1(f_2(x)) = f_1(y) = z = f_1 \circ f_2(x),$ hence $f_1, f_2 \in S \longrightarrow f_1 \circ f_2 \in S.$
ii) $id$ is an isomorphism $ \longrightarrow id \in S$.
iii) $ f $ is an isomorphism $ \longrightarrow \exists f^{-1}$, one can show that $ f^{-1} $ is also an isomorphism, $ \longrightarrow f^{-1} \in S$.
Now, assume that $ | S | = 1$, then I want to show that $ G $ is abelian and each element has an order of 1 or 2.
Im kinda lost with that $ | S | = 1 $. If $ f \in S $ then $f^{-1} \in S $ due to previous result, should not $ | S | = 1 $ imply that $ f = f^{-1} = id$? And how do I move forward from this? Any hints is muy appreciated!