Proof.
We first show that $(1)$ and $(1')$ are equivalent. If $\epsilon $ is an isomorphism, then $f \circ g$ is isomorphic to $\operatorname{id}_{D}$ (as an object of the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}(D,D)$) and is therefore an isomorphism of $\operatorname{\mathcal{C}}$ (Remark 2.2.8.23). Conversely, suppose that $f \circ g$ is an isomorphism. Then it is invertible when viewed as an object of the monoidal category $\underline{\operatorname{End}}_{\operatorname{\mathcal{C}}}(D)$. Since $(\eta , \epsilon )$ is an adjunction, we can regard $f \circ g$ as a coalgebra object of $\underline{\operatorname{End}}_{\operatorname{\mathcal{C}}}(D)$ with counit $\epsilon $ (see Remark 
). Applying Proposition 2.1.5.23 (to the monoidal category $\underline{\operatorname{End}}_{\operatorname{\mathcal{C}}}(D)^{\operatorname{op}}$), we deduce that $\epsilon $ is an isomorphism.
We now show that $(1)$ implies $(2)$. Assume that $\epsilon : f \circ g \Rightarrow \operatorname{id}_{D}$ is an isomorphism. Axiom $(Z1)$ of Definition 6.1.1.1 guarantees that the composition
\[ f \xRightarrow {\sim } f \circ \operatorname{id}_{C} \xRightarrow {\operatorname{id}_ f \circ \eta } f \circ (g \circ f) \xRightarrow {\sim } (f \circ g) \circ f \xRightarrow [\sim ]{\epsilon \circ \operatorname{id}_ f} \operatorname{id}_{D} \circ f \xRightarrow {\sim } f \]
is equal to the identity $2$-morphism $\operatorname{id}_{f}$, which proves that the horizontal composition $\operatorname{id}_{f} \circ \eta $ is an isomorphism in $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(C,D)$. Similarly, it follows from axiom $(Z2)$ of Definition 6.1.1.1 that the horizontal composition $\eta \circ \operatorname{id}_{g}$ is an isomorphism in $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(D,C)$. For every $1$-morphism $d: T \rightarrow D$ in $\operatorname{\mathcal{C}}$, the map
\[ f \circ (g \circ d) \xRightarrow [\sim ]{ \alpha _{f,g,d} } (f \circ g) \circ d \xRightarrow [\sim ]{\epsilon \circ \operatorname{id}_ d} \operatorname{id}_ D \circ d \xRightarrow [\sim ]{ \lambda _ d } d \]
is an isomorphism, so that $d$ belongs to the essential image of the functor $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C) \xrightarrow { f \circ } \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,D)$.
We now show that $(2)$ implies $(2')$. Let $d: T \rightarrow D$ be a $1$-morphism of $\operatorname{\mathcal{C}}$. If the functor $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C) \xrightarrow { f \circ } \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,D)$ is essentially surjective, then $d$ is isomorphic to $f \circ c$ for some $1$-morphism $c: T \rightarrow C$ of $\operatorname{\mathcal{C}}$. If $\operatorname{id}_{f} \circ \eta $ is an isomorphism, then the chain of isomorphisms
\[ f \circ c \xRightarrow {\sim } (f \circ \operatorname{id}_{C}) \circ c \xRightarrow {(\operatorname{id}_ f \circ \eta ) \circ \operatorname{id}_ c} (f \circ (g \circ f) ) \circ c \xRightarrow {\sim } f \circ ( (g \circ f) \circ c) \xRightarrow {\sim } f \circ (g \circ (f \circ c) ) \]
shows that $d$ belongs to the essential image of the composite functor
\[ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T, D) \xrightarrow { g \circ } \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}( T, C) \xrightarrow { f \circ } \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}( T, D ). \]
We next show that $(2')$ implies $(3)$. Fix an object $T \in \operatorname{\mathcal{C}}$ and a pair of $1$-morphisms $d,d': T \rightarrow D$; we wish to show that the composition map
\[ \operatorname{Hom}_{\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,D) }( d', d) \rightarrow \operatorname{Hom}_{\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C) }( g \circ d', g \circ d) \]
is a bijection. By virtue of assumption $(2')$, we may assume that $d' = f \circ c$, where $c: T \rightarrow C$ is a $1$-morphism of the form $g \circ d''$. By virtue of Proposition 6.1.2.9, the composition
\begin{eqnarray*} \operatorname{Hom}_{\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,D)}( f \circ c, d ) & \rightarrow & \operatorname{Hom}_{\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)}( g \circ (f \circ c), g \circ d ) \\ & \simeq & \operatorname{Hom}_{\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)}( (g \circ f) \circ c, g \circ d ) \\ & \xrightarrow { \eta \circ \operatorname{id}_ c } & \operatorname{Hom}_{\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)}( \operatorname{id}_ C \circ c, g \circ d ) \\ & \simeq & \operatorname{Hom}_{\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)}( c, g \circ d ) \end{eqnarray*}
is a bijection. It will therefore suffice to show that the $2$-morphism $(\eta \circ \operatorname{id}_ c): \operatorname{id}_{C} \circ c \Rightarrow (g \circ f) \circ c$ is an isomorphism. This follows from assumption $(2')$, since $(\eta \circ \operatorname{id}_ c)$ can be rewritten as a composition
\[ \operatorname{id}_{C} \circ (g \circ d'') \xRightarrow {\sim } (\operatorname{id}_ C \circ g) \circ d'' \xRightarrow {(\eta \circ \operatorname{id}_ g) \circ \operatorname{id}_{d''} } ((g \circ f) \circ g) \circ d'' \simeq (g \circ f) \circ (g \circ d''). \]
The implication $(3) \Rightarrow (3')$ is clear. We will complete the proof by showing that $(3')$ implies $(1)$. Assume that $(3')$ is satisfied; we wish to show that the $2$-morphism $\epsilon : f \circ g \Rightarrow \operatorname{id}_{D}$ is an isomorphism. To prove this, it will suffice to show that for every $1$-morphism $u: D \rightarrow D$, vertical precomposition with $\epsilon $ induces a bijection $\operatorname{Hom}_{ \underline{\operatorname{End}}_{\operatorname{\mathcal{C}}}(D)}( \operatorname{id}_ D, u ) \rightarrow \operatorname{Hom}_{ \underline{\operatorname{End}}_{\operatorname{\mathcal{C}}}(D)}( f \circ g, u )$. We now observe that this map fits into a commutative diagram
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{ \underline{\operatorname{End}}_{\operatorname{\mathcal{C}}}(D)}( \operatorname{id}_ D, u ) \ar [r]^-{\epsilon } \ar [d]^{\operatorname{id}_ g \circ } & \operatorname{Hom}_{ \underline{\operatorname{End}}_{\operatorname{\mathcal{C}}}(D)}( f \circ g, u ) \ar [d] \\ \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(D,C)}( g \circ \operatorname{id}_ D, g \circ u ) \ar [r]^-{\sim } & \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(D,C)}( g, g \circ u ) } \]
where the bottom horizontal map is induced by the right unit constraint $\rho _{g}: g \circ \operatorname{id}_ D \xRightarrow {\sim } g$, the right vertical map is given by the formation of right adjuncts with respect to $\eta $ (and is therefore bijective by virtue of Corollary 6.1.2.6), and the left vertical map is bijective by virtue of assumption $(3')$.
$\square$