$f(x) = x^6, g(x) = x^{10}$ endomorphisms $\implies G$ is abelian 
Let $(G, \cdot)$ be a group in which the functions $f: G \to G, f(x) = x^6$ and $g : G \to G, g(x) = x^{10}$ are endomorphisms and $f$ is injective. Prove that $G$ is an abelian group.

We need to prove that $f(xy) = f(yx), \forall x,y \in G$.
Because $f$ is an endomorphism, then $(xy)^6 = x^6y^6 \iff (yx)^5 = x^5y^5, \forall x,y \in G$. So, $x^6y^6 = (xy)^6 = (xy)^5(xy) = y^5x^5xy \implies x^6y^5 = y^5x^6, \forall x,y \in G$. Swaping $x$ and $y$, we get that $x^5y^6 = y^6x^5, \forall x,y \in G (*)$.
In the same way we obtain that $x^{10}y^9 = y^9x^{10}, \forall x,y \in G (**)$. Therefore, $x^5y^3 = y^3x^5$ (using $(*)$ and $(**)$), but I don't know how to continue from here.

This is an extended comment rather than an answer.
The assumption that $f$ is injective is necessary.
Indeed, the article Abelian Forcing Sets by Joseph A. Gallian and  Michael Reid contains the following result:

Definition: a set of integers $T$ is called abelian-forcing if for
  any group $G$, if the map $x\mapsto x^t$ is an endomorphism for every
  $t\in T$ then $G$ has to be abelian.
Theorem: a set of integers $T$ is abelian-forcing if and only if the
  gcd of the numbers $t(t-1)$ where $t$ runs over $T$ is equal to $2$.

In the present case, the $\gcd (5\times 6,\  9\times 10)=30\neq 2$.
