Remark

One can prove an analogous statement, assuming $B_n(C')$ and $C'/B_n(C')$ are injective for each $n$.

Set for simplicity $B = B(C)$, $Z = Z(C)$, $H = H(C)$ and $H' = H(C')$. Since $B$ is projective, we derive from (1) an exact sequence

(12) $0 \to \mathrm{Homgr}_A(B, C')(1) \xrightarrow{\mathrm{Homgr}(\delta, 1)} \mathrm{Homgr}_A(C, C')$