Is there a $SL(2,\mathbb{Z})$-action on $\mathbb{Z}$? Is there a $SL(2,\mathbb{Z})$-action on $\mathbb{Z}$?
I read this somewhere without proof and I am not sure if this is true.
Thank you for your help.

(This is completely different to my first 'answer', which was simply wrong.)
Denote by $\text{End}(\mathbb{Z})$ the semi-group of group endomorphisms of $\mathbb{Z}$. Since $\mathbb{Z}$ is cyclic, any endomorphism is determined by the image of the generator $1$, and since $1 \mapsto n$ is an endomorphism for any $n\in\mathbb{Z}$, this is all of them.
Since $SL(2,\mathbb{Z})$ is a group, all of its elements are invertible, so must map to invertible endomorphisms, i.e. automorphisms.  Obviously these are given only by $n = \pm 1$ in the notation above.  So $\text{Aut}(\mathbb{Z}) \cong \mathbb{Z}_2$.  So the question becomes: is there a non-trivial homomorphism $\phi : SL(2,\mathbb{Z}) \to \mathbb{Z}_2$?
Since $\mathbb{Z}_2$ is Abelian, $\phi$ must factor through the Abelianisation of $SL(2,\mathbb{Z})$, which is$^*$ $\mathbb{Z}_{12}$.  There is a unique surjective homomorphism $\mathbb{Z}_{12} \to \mathbb{Z}_2$, and therefore a unique surjective $\phi$, which gives a unique non-trivial action of $SL(2,\mathbb{Z})$ on $\mathbb{Z}$.  Unfortunately, I can't see an easy way to decide whether a given $SL(2,\mathbb{Z})$ matrix maps to $1$ or $-1$, but maybe somebody else can.
$^*$A proof of this can be found at the link provided in the comments.
