Commuting right adjoints implies commuting left adjoints. Say that I have a diagram of right adjoints 
$$
\require{AMScd}
\begin{CD}
A @<{R_1}<< B \\
@V{R_3}VV @VV{R_2}V \\
D @<<{R_4}< C
\end{CD}
$$
does it follow that the diagram of left adjoints
$$
\require{AMScd}
\begin{CD}
A @>{L_1}>> B \\
@A{L_3}AA @A{L_2}AA \\
D @>>{L_4}> C
\end{CD}
$$
commutes also, where $L_1$ is left adjoint to $R_1$ and so on. 
It seems like something that should be possible (it works for every example I've written down but doesn't seem to rely on the right adjoints) but I'm not convinced. 
 A: There are two results that piece together to give you your answer.

Theorem 1. If $\mathcal{C} \overset{F}{\underset{G}{\rightleftarrows}} \mathcal{D} \overset{K}{\underset{H}{\rightleftarrows}} \mathcal{E}$ are functors with $F \dashv G$ and $K \dashv H$, then $G \circ H \dashv K \circ F$.
Theorem 2. If $F \dashv G_1$ and $F \dashv G_2$, then $G_1 \cong G_2$; and if $F_1 \dashv G$ and $F_2 \dashv G$, then $F_1 \cong F_2$.

So assume that $R_3 \circ R_1 = R_4 \circ R_2$. Then $L_1 \circ L_3 \dashv R_3 \circ R_1$ by Theorem 1, and $L_2 \circ L_4 \dashv R_4 \circ R_2 = R_3 \circ R_1$ by Theorem 1, and so $L_1 \circ L_3 \cong L_2 \circ L_4$ by Theorem 2.
Hence the square of left adjoints commutes up to natural isomorphism.
Unfortunately you cannot expect to obtain a strong result than commutativity up to natural isomorphism, since adjoints themselves are only defined up to natural isomorphism.
Here's an example. The following square certainly commutes, where $\Delta : \mathbf{Set} \to \mathbf{Set} \times \mathbf{Set}$ is the diagonal functor, defined on objects by $A \mapsto (A,A)$.
$$\require{amsCD}
\begin{CD}
\mathbf{Set} \times \mathbf{Set} @<{\Delta}<< \mathbf{Set} \\
@V{\Delta \times \mathrm{id}}VV @VV{\Delta}V \\
\mathbf{Set} \times \mathbf{Set} \times \mathbf{Set} @<<{\Delta \times \mathrm{id}}< \mathbf{Set} \times \mathbf{Set}
\end{CD}$$
Let ${+} : \mathbf{Set} \times \mathbf{Set} \to \mathbf{Set}$ be the coproduct functor defined on objects by
$${+}(A,B) \mapsto A+B = (A \times \{ 0 \}) \cup (B \times \{ 1 \})$$
and let $+' : \mathbf{Set} \times \mathbf{Set} \to \mathbf{Set}$ be the coproduct functor defined on objects by
$${+'}(A,B) \mapsto B+A = (A \times \{ 1 \}) \cup (B \times \{ 0 \})$$
Then ${+} \dashv \Delta$ and ${+'} \dashv \Delta$, and ${+} \times \mathrm{id} \dashv \Delta \times \mathrm{id}$ and ${+'} \times \mathrm{id} \dashv \Delta \times \mathrm{id}$, and so we obtain the following diagram of left adjoints
$$\require{AMScd}
\begin{CD}
\mathbf{Set} \times \mathbf{Set} @>{+}>> \mathbf{Set} \\
@A{{+} \times \mathrm{id}}AA @AA{+'}A \\
\mathbf{Set} \times \mathbf{Set} \times \mathbf{Set} @>>{{+'} \times \mathrm{id}}> \mathbf{Set} \times \mathbf{Set}
\end{CD}$$
This commutes up to natural isomorphism, since $(A+B)+C \cong (A+'B)+'C$ naturally in $A,B,C$, but it does not commute strictly.
A: Clive Newstead has already given an excellent answer (which I've upvoted), but I wanted to add another perspective.
The goal is to show that $L_1(L_3(x))\simeq L_2(L_4(x))$ for all $x\in D$, where $\simeq$ denotes natural isomorphism. Then by the Yoneda embedding it is equivalent to show that
$$\newcommand\Hom{\operatorname{Hom}}\Hom(L_1 L_3,-)\simeq \Hom(L_2 L_4,-),$$
however this follows immediately from the adjunctions and the fact that the right adjoints commute:
$$\Hom(L_1 L_3,-)\simeq \Hom(L_3,R_1)\simeq \Hom(-,R_3 R_1)\simeq \Hom(-,R_4 R_2)\simeq \Hom(L_2 L_4,-).$$
This is essentially equivalent to just applying the results that Clive cited, but when working with adjoints, I often find it illuminating to spell out the actual sequence of natural isomorphisms of Hom sets.
