This is in fact related to this question. Let $\phi$ be an injective but not surjective homomorphism $G\to G$. Then under very mild conditions (see below) the powers $\phi, \phi^2,...,\phi^n,...$ are not conjugate in $G$ for all $n$ (see the definition in the linked question). Indeed, suppose that there exists a $g$ such that $\phi^{i+j}(x)=g\phi^i(x)g^{-1}$ for every $x\in G$ for some $i,j>0$. Then applying $\phi$, we get $\phi^{i+j+1}(x) = \phi(g)\phi^{i+1}(x)\phi(g)^{-1} = g\phi^{i+1}(x)g^{-1}$. Therefore $a=g^{-1}\phi(g)$ commutes with $\phi^{i+1}(x)$ for every $x$. Suppose that centralizers of subgroups of $G$ that are isomorphic to $G$ are trivial (it is the mild condition I mentioned before, that holds for non-elementary hyperbolic, relatively hyperbolic, etc. groups). Then $a=1$, hence $g=\phi(g)$. But then $\phi^{i+j}(x)=\phi^i(g)\phi^i(x)(\phi^i(g))^{-1}=\phi^i(gxg^{-1})$ for every $x$. Hence $\phi^j(x)=gxg^{-1}$ for every $x$ and $\phi^j$ is surjective, a contradiction.

This implies (see the question above again) that $G$ acts non-trivially on an asymptotic cone of itself. Thus, for example, if $G$ is hyperbolic with property (T), it is co-Hopfian because its asymptotic cones are $\mathbb{R}$-trees and a group with property (T) cannot act on an $\mathbb{R}$-tree non-trivially. Much stronger results can be found in Sela, Z.
Endomorphisms of hyperbolic groups. I. The Hopf property.
Topology 38 (1999), no. 2, 301–321 and Druţu, Cornelia, Sapir, Mark V.
Groups acting on tree-graded spaces and splittings of relatively hyperbolic groups.
Adv. Math. 217 (2008), no. 3, 1313–1367.