$f(x) = x^6, g(x) = x^{10}$ 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.

The hypotheses are satisfied in any group of exponent $5$ (because $f(x)=x$ and $g(x)=1$ for all $x \in G$), and there are nonabelian groups of exponent $5$, so the claimed result is not correct.
Perhaps you should be assuming that both $f$ and $g$ are injective.
