My question is how can we see that if $F$ and $F'$ are both left adjoints of $G$, there is natural isomorphism between $F$ and $F'$? So how can we show that adjoints are unique upto isomorphism?
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If $F \vdash G$, $F' \vdash G$, then we have units/counits $\epsilon:I \rightarrow FG, \eta:GF \rightarrow I$, and $\epsilon':I \rightarrow F'G, \eta':GF' \rightarrow I$. Then $\epsilon F':F' \rightarrow FGF'$, while $F\eta' :FGF' \rightarrow F$. Composing these gives a natural transformation $F \rightarrow F'$. Now you, of course, has to find the inverse map and show it actually is an inverse. The nice thing about this approach is it's already natural! For me, the key is to understand how to translate your information into the "algebra of functors" and work with it there.
We have $\hom(Fx,y) \cong \hom(x,Gy) \cong \hom(F'x,y)$ natural in $x,y$, hence $F \cong F'$ (Yoneda).
PS: Compare this with the proof that adjoint operators between Hilbert spaces are unique.