Ah, sorry, I over-interpreted your statement. If $G$ does not map onto $\mathbb{Z}$ then these statements are essentially equivalent, which is what I thought you were saying (that is, if $G$ does not surject onto $\mathbb{Z}$ then $G\rtimes_{\psi_1}\mathbb{Z}\cong G\rtimes_{\psi_2}\mathbb{Z}$ if and only if $[\psi_1]$ is conjugate to $[\psi_2]$ or to $[\psi_2^{-1}]$ modulo the inner automorphisms (by $[\psi]$ I mean the outer automorphism $\psi\operatorname{Inn}(G)$).)

Does your statement about the isomorphism class depending only on the conjugacy class of $\psi$ require that $G$ does not surject onto $\mathbb{Z}$?

It is worth pointing out that Bumagin-Wise (Every group is the outer automorphism group of a finitely generated group, J. Pure App. Alg. 2007) used a variation on Rips construction to prove that given any finitely presented group $Q$ there exists a finitely generated, residually finite group $Q$ such that $\operatorname{Out}(G_Q)\cong Q$. Here, $G_Q$ is the kernel of Rips construction, so if we take $Q$ to be trivial then $G_Q$ is $C{\prime}(1/6)$ and we get an explicit presentation. If $Q$ is finite then $G_Q$ is still hyperbolic, large, virtually compact special, etc.

Oh, that is excellent. I will look forward to it!

I am struggling to find where you prove that the centralisers are preserved in your "aspherical groups..." paper. Could you perhaps point me in the right direction? Also, having come back to this after a while I have realised that the two constructions you give only work for finitely-generated groups. Do you know if it is "not impossible" for arbitrary recursively presented groups? (I suppose I am asking: where does the limitation of finitely generated come from? Is it just the techniques, or is it more fundamental than that?)

If one of your relators is a proper power, $r_i=s_i^n$ and $n>1$, then $f=n-1$ works.

Sorry, you are right, I was assuming the groups $G$ were finitely generated. I have edited the question accordingly.