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From the wikipedia page on the Equivalence of Categories:

If $F : C \rightarrow D$ is an equivalence, [then] the object $c$ of $C$ is an initial object if and only if $Fc$ is an initial object of $D$

Proposed Counter-Example:

Let $C = \mathbf{1}$, with two objects $a$ and $b$ and one non-identity arrow $f : a \rightarrow b$.

Let $D$ have three objects: $a$, $b$, and $c$. Let there be one non-identity arrow from $a$ to $b$ also called $f : a \rightarrow b$. Let there be an invertible arrow $g : b \rightarrow c$ with $g^{-1} : c \rightarrow b$. Then let $h: b \rightarrow c$.

Finally, let $F$ be the identity functor on $C$, so that clearly $F$ is fully faithful and dense (and hence $C$ and $D$ are equivalent).

Then $a$ is an initial object in $C$, but $Fa$ is not an initial object in $D$, since $g \circ f : a \rightarrow c \ne h \circ f : a \rightarrow c$.

Question: I'm sure I'm getting something wrong, but it's not clear to me what it is. Why is the preceding not a true counter-example?

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    You haven't specified $D$ fully. How does $h$ compose with $f$ and $g$? Once you specify this I think you'll find that either $Fa$ becomes initial or $F$ fails to be an equivalence. – 2017-02-20
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    @QiaochuYuan: What do you mean? Mustn't $h \circ f$ exist by the composition axiom for categories? And similarly, $g^{-1} \circ h$ must also be an arrow. So what is left to specify? – 2017-02-20
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    So, you might be freely adjoining $h$, and not imposing any further equalities than those forced by the category theory axioms. In that case $F$ fails to be an equivalence: it is no longer fully faithful. – 2017-02-20
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    (Specifically, $F(b)$ has two endomorphisms.) – 2017-02-20

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