Suppose we have a diagram 
$\require{AMScd}$
\begin{CD}
A @>a>> B\\
@V b V V @VV c V\\
C @>>d> D @>e>> E \\
@VfVV @VVgV @VVhV  \\
F @>>i> G  @>>j> H
\end{CD}
of schemes (or could be topological spaces instead), and we have an $\ell$-adic sheaf $\mathcal{K}$ on $H$ (or could be a constructible sheaf on a topological sheaf). 

Assume the following squares are (commutative and) Cartesian: 
$\require{AMScd}$
\begin{CD}
A @>a>> B\\
@V b V V @VV c V\\
C @>>d> D 
\end{CD}\begin{CD}
D @>a>> E\\
@V b V V @VV c V\\
G @>>d> H
\end{CD}

$\require{AMScd}$
\begin{CD}
A @>>d> B\\
@Vf \circ b VV @VVg \circ c V \\
F @>>i> G
\end{CD} 


$\require{AMScd}$
\begin{CD}
C @>>e \circ d> E\\
@Vf \circ b VV @VVg \circ c V \\
F @>>j \circ i> G
\end{CD} 

but 
$\require{AMScd}$
\begin{CD}
C @>>d> D\\
@VfVV @VVhV \\
F @>>i> G
\end{CD} 
is **not even commutative.** In any Cartesian square, there is a base change morphism of $*$ and $!$-pullbacks, for example, $a^* c^! \rightarrow b^! d^*$. We consider pulling back $\mathcal{K}$ from $H$ to $A$, using $!$-pullback along the vertical morphisms and $*$-pullback along the horizontal morphisms. This yields two morphisms with source the sheaf obtained by pulling back along the outer right path from $H$ to $A$, and target the sheaf obtained by pulling back along the bottom left path from $H$ to $A$: 

$$a^* c^! e^* h^! \mathcal{K} \rightarrow b^! f^! i^* j^* \mathcal{K}$$ 

1. The first comes from composing the base change maps obtained by cutting up the diagram into two Cartesian squares as below 

$\require{AMScd}$
\begin{CD}
A @>a>> B\\
@V b V V @VV c V\\
C @>>d> D @>e>> E \\
@VfVV @. @VVhV  \\
F @>>i> G  @>>j> H
\end{CD}

2. The second comes from composing the base change maps obtained by cutting up the diagram into two Cartesian squares as below 
$\require{AMScd}$
\begin{CD}
A @>a>> B\\
@V b V V @VV c V\\
C @. D @>e>> E \\
@VfVV @VVgV @VVhV  \\
F @>>i> G  @>>j> H
\end{CD}

**Question:** must the two morphisms $a^* c^! e^* h^! \mathcal{K} \rightarrow b^! f^! i^* j^* \mathcal{K}$ agree? 

One usually assumes heuristically that two canonical morphisms between the same source and target agree, but in this case I feel unsure. If the non-commutative diagram were commutative (hence automatically Cartesian), then it would be immediate. 

Finally, an example of this situation: 

$\require{AMScd}$
\begin{CD}
A @>a>> B\\
@V b V V @VV c = (pt, c_2) V\\
C @>>d = \Delta> C \times C @>e = pr_1>> C \\
@VfVV @VVg = pr_2 V @VVhV  \\
pt @>>i> C  @>>j> pt
\end{CD}

I would be happy with a proof or counterexample even just for this example.