I was trying to solve the following functional equation:
$$f\big(f(x-y)\big)=f(x)-f(y)\text.$$
And I concluded that $f$ must be additive and bijective. The question is:
Let $f:\mathbb{R} \to \mathbb{R}$ be an additive bijective function. Is it true that $f$ has to be of the form $f(x)=cx$ for some $c\in\mathbb{R}\setminus\{0\}$?
What if we impose the condition that $f\big(f(x)\big)=f(x)$. Is the identity map the only solution?
Edit: if $f$ is surjective and $f\big(f(x)\big)=f(x)$, then $f$ is the identity map (almost by definition).
While there are uncountably many wild additive bijections on $ \mathbb R $ (as @MarcoVergamini has indicated in a comment above), the original equation you encountered has only one injective/surjective solution: the identity function. It's straightforward to check that the identity function is a bijective solution. To see that it's the only injective/surjective solution, let $ x = y = 0 $ in $$ f \big( f ( x - y ) \big) = f ( x ) - f ( y ) \tag { * } \label 0 $$ to get $ f \big( f ( 0 ) \big) = 0 $. Then, put $ x = f ( 0 ) $ and $ y = 0 $ in \eqref{0} to see that $ f ( 0 ) = 0 $. By setting $ y = 0 $ in \eqref{0} you have $ f \big( f ( x ) \big) = f ( x ) $. This means $ f ( x ) = x $ for every $ x $ in the range of $ f $. So, if $ f $ is supposed to be surjective, it follows that $ f $ must be the identity function. Also, if $ f $ is supposed to be injective, $ f \big( f ( x ) \big) = f ( x ) $ directly implies $ f ( x ) = x $ for all $ x $.
But there seems to be a problem with your claim that solutions of \eqref{0} must be bijective. Just similar to the construction of wild additive bijections, you can use a Hamel basis $ \mathcal H $ of $ \mathbb R $ over $ \mathbb Q $ to construct wild solutions of \eqref{0}: choose a nonempty subset $ \mathcal I $ of $ \mathcal H $, map every member of $ \mathcal I $ to itself and every member of $ \mathcal H \setminus \mathcal I $ to an arbitrary member of $ \mathcal I $, and finally extend this mapping linearly (over $ \mathbb Q $) to a function $ f : \mathbb R \to \mathbb R $. It's straightforward to check that this $ f $ will satisfy \eqref{0}. Therefore, unless any further restriction on $ f $ is given, it cannot be concluded solely from \eqref{0} that $ f $ must be injective/surjective.
But additivity of $ f $ is secured by \eqref{0}. $ f \big( f ( x ) \big) = f ( x ) $ can be used to rewrite \eqref{0} as $ f ( x - y ) = f ( x ) - f ( y ) $. Substituting $ x + y $ for $ x $ and rearranging the terms, that equation is transformed to the Cauchy functional equation. So, this part of your claim is definitely true.