I've been trying to prove the following claim, but am now unsure about its truth. Is it true, and if so, where can I find a proof? 


**Claim:** For any categories C, D, E such that C and D are equivalent, 


(i) The set $Hom($C, E$)$ of functors from C to E is in bijective correspondence with the set $Hom($D, E$)$ of functors from D to E, i.e. $Hom($C, E$)$ $\simeq$ $Hom($D, E$)$. 


(ii) The set $Hom($E, C$)$ of functors from E to C is in bijective correspondence with the set $Hom($E, D$)$ of functors from E to D, i.e. $Hom($E, C$)$ $\simeq$ $Hom($E, D$)$. 



Bijection between hom sets of equivalent categories?
I will turn my comment into an answer.
If $X$ is any nonempty set, then let ${\mathcal C}\_X$ be the category
which has $X$ as its class of objects and, for each pair
$(x\_1,x\_2)\in X^2$, has exactly one morphism
$\varphi\_{x\_1,x\_2}:x\_1\to x\_2$. Every morphism
in ${\mathcal C}\_X$ is necessarily an isomorphism.


Any function $f: X\to Y$ is the object part of a unique
functor $F: {\mathcal C}\_X\to {\mathcal C}\_Y$,
and any two such functors are naturally isomorphic.
In particular, ${\mathcal C}\_X$ and
${\mathcal C}\_Y$ are categorically equivalent
for any nonempty $X$ and $Y$.


Hence for $X = \{0\}$, $Y = \{0,1\}$, and
${\sf C} = {\mathcal C}\_X$, 
${\sf D} = {\sf E} = {\mathcal C}\_Y$ we have  

(i) $|\textrm{Hom}({\sf C}, {\sf E})| = 2$ and 
$|\textrm{Hom}({\sf D}, {\sf E})| = 4$, while   

(ii) $|\textrm{Hom}({\sf E}, {\sf C})| = 1$ and 
$|\textrm{Hom}({\sf E}, {\sf D})| = 4$.




