I was at a class, when the teacher used the following statement to prove some facts: $G$ is a group and $f \in End(G)$ such that $f(g)$ is conjugate to $g$ for all $g \in G$ then $f \in Aut(G)$.
Actually, in the end he proved something more about the function that showed that $f$ was in $Aut(G)$ (he proved that $f = id_G$).
But, anyways, I was curious if this was true, or if he made a mistake, I tried something: Firstly, we know that $f$ may be injective, because if $g \in ker(f)$ then $1 =f(g)$ is conjugate to $g$ then $g=1$, so $ker(f) = 1$.
Then I proved something else:
Firstly if we consider $f: G \rightarrow Imf$, we may have that $f$ is an isomorphism. It's true, since $f$ in injective as we proved, and surjective by the restriction of counter-domain.
And actually, if we consider $f|_{Im(f^n)} : Im(f^n) \rightarrow Im(f^{n+1})$ would be an isomorphism by the same argument.
Then I tried to obtain some information about the orbits. I did this by proving the following statement about $f$:
$\textbf{Statement}$: Let $g \in G$, and $O(g)$ the orbit or conjugacy class of $g$. If $O(g)$ is finite then $g \in Im(f)$ and exists some $i>0$ such that $f^i(g) = g$
$\textbf{Proof}$: Since $O(g)$ is finite and $f^n(g) \in O(g)$ for all $n$ then we must have $i,j$, $i<j$ such that $f^i(g) = f^j(g)$. Then, we have $f^i(g) = f^i(f^{j-i}(g)) \Rightarrow g = f^{j-i}(g)$ because $f^i$ will be injective as is compose of injective functions. Since $i<j$ we must have $i \leq j-1$ then $j-i-1 = j-1-i \geq 0$, so $f(f^{j-i-1}(g)) = g$ then $g \in Im(f)$
So if $f$ is not surjective then we must have an infinity conjugacy class.
I also proved a corollary of the previous statement that is: if $g \not \in Im(f)$ then there isn't $a_1, ... ,a_k \in G$ all with finite conjugacy class such that $ g= a_1 \cdot ... \cdot a_k$. It's true, otherwise it'd exist $i_1,... i_k$ such that $f^{i_l}(a_l) = a_l$. Then we would have $f^{i_1 \cdot ... \cdot i_k}(g) = f^{i_1 \cdot ... \cdot i_k}(a_1 ... a_k) = f^{i_1 \cdot ... \cdot i_k}(a_1) \cdot ... \cdot f^{i_1 \cdot ... \cdot i_k}(a_k) = a_1 \cdot ... \cdot a_k = g$ and this would imply that $g \in Im(f)$.
Then corollary of corollary: if we multiply g in the left or in the right by elements whose inverse has finite conjugacy class then it may not have finite conjugacy class: for example if $a^{-1}$ has finite conjugacy class then if $ag$ had finite conjugacy class then, we would have a contradiction, since $g = a^{-1}(ag)$
I also tried something involving the left-inverse but I couldn't achieve any further.
I'd like to know if this is true or has some counter-examples.
As I think you pointed out, this is true for finite groups $G$. But it's not true in general.
Let $X$ be the symmetric group on the set ${\mathbb Z}$ of integers, and let $Y$ be the normal subgroup consisting of the finitary permutations; that is, those with finite support.
Let $G$ be the subgroup of $Y$ that stabilizes all negative integers; that is permutations of ${\mathbb Z}$ with finite support contained in the set $\{ n : n \in {\mathbb Z} \mid n \ge 0 \}$, and let $g \in X$ be defined by $g: n \mapsto n+1\ \forall n \in {\mathbb Z}$.
Now if we define $f:G \to G$ as conjugation by $g$, that is $f: x \mapsto gxg^{-1}$, then $f$ maps $G$ to the stabilizer in $G$ of $0$, so $f$ is an injective but not surjective endomorphism of $G$. Furthermore, since $x$ and $f(x)$ both have finite support, they are conjugate in $G$ for all $x \in G$.