Here is a very silly question:
Adjoint functors satisfy
$$\mathrm{hom}_{\mathcal{C}}(FA,B) \cong \mathrm{hom}_{\mathcal{D}}(A,GB).$$
I consider numbers $a,b$ and read this as
$$b^{\,f(a)}=g(b)^a.$$
If the objects in the categories can be assigned cardinalities, do the functors actually fulfill a relation along those lines?
$\bf Edit$: If e.g. $|B^{FA}|=|B|^{|FA|}$ does make sense, just taking the cardinalities of the hom-sets tells us
$$\frac{|FA|}{|A|}=\log_{|B|}|GB|.$$
E.g. in a category with object being sets, the adjoint functors $FA:=A\times I$ and $GB:=B^I$ have
$\frac{|FA|}{|A|}=\frac{|A\times I|}{|A|}=|I|\ \ \ $ and $\ \ \ \log_{|B|}|GB|=\log_{|B|}|B^I|=|I|$.
I don't know if we can interpret this as an equation of numbers, but what about the following: For objects $a,b$ of a category the exponential $b^a$ is an object with a natural bijection $\hom(c,b^a) \cong \hom(c \times a,b)$. A category is called cartesian closed if it has all binary products and exponentials for any two objects. Now assume that $C$ is cartesian closed. Assume that $F : C \to C$ is left adjoint to $G$. We want to compare $b^{F(a)}$ with $G(b)^a$. Thus, we want to compare $$\hom(c,b^{F(a)})=\hom(c \times F(a),b)$$ with $$\hom(c,G(b)^a) = \hom(c \times a,G(b)) = \hom(F(c \times a),b)$$ Therefore, a natural isomorphism $b^{F(a)} \cong G(b)^a$ corresponds to a natural isomorphism $c \times F(a) \cong F(c \times a)$. Taking $a$ to be the terminal object $1$ and letting $x=F(1)$, we get that $F(c) \cong c \times x$. It follows $G(b) \cong b^x$. Therefore, the isomorphism reduces to the plausible isomorphism $$b^{a \times x} \cong (b^x)^a.$$ This can be also checked directly. For arbitrary $F$, I don't know how to make sense of $b^{F(a)} \cong G(b)^a$.