Remark 4.1.4 in Riehl's, *Category Theory in Context* 
I'm having trouble proving the statement in Remark 4.1.4, i.e using $k\circ f=g\circ Fh\iff Gk\circ f^\sharp=g^\sharp\circ h$ to prove $Gk\circ f^\sharp =(k\circ f)^\sharp,\; g^\sharp \circ h=(g\circ Fh)^\sharp$. How can I do this?
 A: The point is very simple, if the square with say $F$ commutes, you apply $G$ on the whole diagram and then use the naturality to get rid of the $GFc$ and $GFc'$ and replace them by $c$ and $c'$. Have you tried that? What happened? (Did you see the units and counits appear?) I would draw the thing in diagrams but sadly I don't know how to implement this in MathJax (is it even possible? I see you did this with pictures). 
If you try doing this with equations, the statement that the first diagram commutes is that $g \circ Fh = k \circ f$, and the naturality statement says that if the natural isomorphism is denoted by $\varphi_{c,d} : \mathrm{Hom}(Fc,d) \to \mathrm{Hom}(c,Gd)$, then for any morphisms $\psi : Fc \to d$, $\alpha_c : c'' \to c$ and $\alpha_d : d \to d''$, we have 
$$
\varphi_{c'',d''}(\alpha_d \circ \psi \circ F \alpha_c) = G \alpha_d \circ \varphi_{c,d}(\psi) \circ \alpha_c. 
$$
Applying $\varphi_{c,d'}$ to $g \circ Fh = k \circ f$ (choose the appropriate $c,c'',d,d''$ and morphisms, and plug in the above to get the first and third equality), we get 
$$
 Gk \circ \varphi_{c',d}(f) = \varphi_{c,d'}(k \circ f) = \varphi_{c,d'}(g \circ Fh) = \varphi_{c',d}(g) \circ h.
$$
I am guessing your $^{\sharp}$ notation refers to the natural isomorphism, so we are done. 
(The other direction is completely symmetric.)
Added : to prove what you actually wanted to show, if you pick $h$ to be the identity, then the first and second square become triangles ; the fact that the first triangle commutes implies that the second triangle commutes translates to 
$$
\varphi_{c,d'}(k \circ f) = \varphi_{c,d'}(g) = Gk \circ \varphi_{c,d}(f),
$$
which is exactly what you wanted to prove. Taking $k$ to be the identity instead, you get
$$
\varphi_{c,d'}^{-1}(g^{\sharp} \circ h) = \varphi_{c,d'}^{-1}(f^{\sharp}) = \varphi_{c,d'}^{-1}(g^{\sharp}) \circ Fh = g \circ Fh,
$$
which is again exactly what you wanted to show. 
Hope that helps,
A: This might get you started: 
We want to show
$$\text{Naturality of isomorphisms} \color{red}{\iff} \left( \text{first square commutes} \color{blue}{\iff} \text{second square commutes} \right).$$
($\color{red}{\implies}$) (Only showing the $\color{red}\implies$ direction for now)
($\color{blue}{\implies}$) Suppose $k \circ f = g \circ Hf$. Then $(k \circ f)^\sharp = (g \circ Hf)^\sharp$. The former is $Gk \circ f^\sharp$ and the latter is $g^\sharp \circ h$. (To show this, draw squares for the naturality of the bijections in the first variable and in the second variable. Put $g$ in the first diagram and $f$ in the second, and see where they go. Remember that the bijections we're talking about are $l \mapsto l^\sharp$)
($\color{blue}{\impliedby}$) Similar.
