Let $F\colon\mathsf{C}\to\mathsf{D}, G\colon\mathsf{D}\to\mathsf{C}$ together with $\eta\colon 1_{\mathsf{C}}\to \mathsf{GF}, \epsilon\colon\mathsf{FG}\to 1_{\mathsf{D}}$ be an equivalence of categories. I know how to replace $\epsilon$ with $\epsilon'$ and such $\epsilon'$ is still an natural isomorphism and together with $\eta$ they form a unit-counit pair.
Howerver, for some reason I can't figure out with what we should replace $\eta$ if we want that instead. Riehl in the book Category Theory in Context defines $\gamma = G(\epsilon)\circ \eta G$ and claims that
Redefining either $\epsilon$ or $\eta$ so as to absorb the isomorphism $\gamma$ - it will not matter which -we will show that there sulting pair of natural isomorphisms define the unit and the counit of an adjunction $F\dashv G$.
Then she replaces $\epsilon$ with $\epsilon' = \epsilon\circ F(\gamma^{-1})$ and proves that $\eta$ and $\epsilon'$ satisfy the triangle identites for the unit and the counit. I tried replacing $\gamma$ with $\epsilon F\circ F(\eta)$ to no avail.
You don't need to replace $\gamma$.
Set $\ \eta':=\gamma^{-1}F\circ\eta$.
Then we have $G(\varepsilon)\circ \eta'G\ =\ G(\varepsilon)\circ \eta^{-1}GFG \circ G\varepsilon^{-1}FG\circ\eta G$. $$\matrix{GFGx &\overset{\eta GFG}\to& GFGFGx \\ {\scriptstyle G\varepsilon}\downarrow \phantom{ {}_{G\varepsilon}} && \phantom{{}_{G\varepsilon FG}} \downarrow {\scriptstyle G\varepsilon FG}\\ Gx &\underset{\eta G}\to& GFGx}$$ Show that this square is commutative, and as each arrow is invertible, the above composition is $1_{Gx}$.
Similarly, $\varepsilon F\circ F(\eta')\ =\ \varepsilon F\circ F\eta^{-1}GF\circ FG\varepsilon^{-1}F\circ F\eta\ =\ 1_{Fx} $, belonging to the commutativity of $$\matrix{FGFx &\overset{F \eta GF}\to& FGFGFx \\ {\scriptstyle \varepsilon F}\downarrow \phantom{ {}_{\varepsilon F}} && \phantom{{}_{FG\varepsilon F}} \downarrow {\scriptstyle FG\varepsilon F}\\ Fx &\underset{F\eta}\to& FGFx}$$