= Solution
\b[A is the N-bound linkage isomer], $\mathrm{[(NH_3)_5Co(NCS)]^{2+}}$: five ammines plus NCS account for $\mathrm{CoCH_{15}N_6S}$.
The natural route is <conjugate-base substitution>. <Hydroxide> reversibly removes an ammine proton:
$$
\mathrm{[(NH_3)_5CoX]^{2+}+OH^-
\rightleftharpoons[(NH_3)_4(NH_2)CoX]^++H_2O}.
$$
The amido donor promotes loss of X$^-$, producing a short-lived five-coordinate $\mathrm{[(NH_3)_4(NH_2)Co]^{2+}}$ intermediate. That leaving-group-sensitive loss is the slower stage. Subsequent trapping/proton transfers are fast and no longer depend on which halide left.
<Water> capture followed by proton transfer gives the hydroxo complex. Competing SCN$^-$ capture can occur at either end of this <ambidentate ligand>, producing Co-SCN and Co-NCS. Any initially captured S-bound material can undergo the base-assisted linkage rearrangement considered next. <Water> is overwhelmingly abundant, so the stable N-bound product can remain a minor overall product. The slow X-dependent and fast X-independent stages fit loss of X followed by chemistry of a common intermediate; a single rate-limiting direct attack by thiocyanate would not explain that separation naturally.
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