Csdrhrt

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.

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 $xin D$, where $simeq$ denotes natural isomorphism. Then by the Yoneda embedding it is equivalent to show that
$$newcommandHom{operatorname{Hom}}Hom(L_1 L_3,-)simeq Hom(L_2 L_4,-),$$
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.

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.