Let $G$ a cyclic group of order 25. Prove $\forall \phi \in{\rm Aut}(G):|\phi| \neq 3$.
I am a bit stuck and I'm not sure I'm even going in the right path.
Assume with contradiction $\exists \phi \in G$ s.t $|\phi| = 3$. This implies $\forall g \in G$ we have $\phi^3(g) = g$.
From datum $G$ is cyclic consider $g \in G$ s.t $G=< \langle g\rangle$. This implies $\phi(g) = g^\alpha$ for some $\alpha$.
Now we have
$$\begin{align} e_G &= \phi^3(g) \\ &=\phi^2(g^\alpha) \\ &=\phi(\phi(g) ... \phi(g)) \\ &= \phi(g^\alpha ... g^\alpha) \\ &= \phi(g^{\alpha^2})\\ & = g^{\alpha^3}, \end{align}$$
which implies $g = g^{\alpha^3}$.
Now since $G=\langle g\rangle \implies |g| = 25$ and this is where I'm hoping to find contradiction without success.
The automorphism group of a group $G$ has function composition as operation and the identity element is the identity function.
The composition of two automorphisms is again an automorphism, so $\phi^3$ cannot send every element to $e_G$, because this map is not bijective.
So your contradiction assumption should be $\phi^3(x)=x$ for every $x\in G$, with $\phi$ not the identity map. In particular, $\phi^3(g)=g$ where $g$ is a generator of the cyclic group.
Now you know that $\phi(g)=g^a$ for some $a$ with $0<a<25$, so $$ \phi^3(g)=g^{a^3} $$ and therefore $g^{a^3-1}=e_G$. Thus $a^3-1\equiv0\pmod{25}$. This is impossible (prove it).
It's simpler if you consider $G=\mathbb{Z}/n\mathbb{Z}$ and note that the endomorphisms of $G$ have the form $x\mapsto kx$, where $k$ is any integer in the range $0..24$. For what $k$ is this map bijective? Exactly when $kx\equiv1\pmod{25}$ for some $x$, which is the same as saying that $k$ is coprime to $25$. There are $20$ such elements, so the automorphism group has order $20$.