Solve on $\mathbb{R}$:

$$
f(2 f(x)+f(y))=2 x+f(y)
$$
Let $f$ be a potential solution. We will show that $f$ is injective: let $a, b$ be real numbers such that $f(a)=f(b)$. Then $f(2 f(b)+f(b))=f(2 f(a)+f(a))$ hence $2 b+f(b)=$ $2 a+f(a)$ and finally $a=b$. Therefore, $f$ is injective.

By taking $x=0$ we have $f(2 f(0)+f(y))=f(y)$, and with the injectivity of $f$, $f(y)=y-2 f(0)$. We set $c=-2 f(0)$ and with the initial equation for all real $x, y$:

$$
(2(x+c)+(y+c))+c=2 x+(y+c)
$$

thus $c=0$ and the identity is the only solution (this is easily verified).