[proofplan]
We use the defining consequence of an equivalence of categories: the functor $F$ is full and faithful, so for every pair of objects it induces a bijection on morphism sets. Functoriality immediately sends inverse morphisms in $\mathcal C$ to inverse morphisms in $\mathcal D$. Conversely, fullness lifts an isomorphism between $F(X)$ and $F(Y)$ to morphisms between $X$ and $Y$, and faithfulness forces the lifted composites to be identity morphisms. The automorphism statement follows by restricting the endomorphism-set bijection to invertible morphisms and checking preservation of composition.
[/proofplan]
[step:Use functoriality to send isomorphisms forward]
Assume $X,Y\in\mathcal C$ and $X\cong Y$ in $\mathcal C$. Choose morphisms
\begin{align*}
f:X\to Y,
\qquad
g:Y\to X
\end{align*}
such that
\begin{align*}
g\circ f=\operatorname{id}_X,
\qquad
f\circ g=\operatorname{id}_Y.
\end{align*}
Since $F:\mathcal C\to\mathcal D$ is a functor, it preserves identity morphisms and composition. Hence
\begin{align*}
F(g)\circ F(f)
&=F(g\circ f)
=F(\operatorname{id}_X)
=\operatorname{id}_{F(X)},\\
F(f)\circ F(g)
&=F(f\circ g)
=F(\operatorname{id}_Y)
=\operatorname{id}_{F(Y)}.
\end{align*}
Therefore $F(f):F(X)\to F(Y)$ is an isomorphism in $\mathcal D$, with inverse $F(g):F(Y)\to F(X)$. Thus $F(X)\cong F(Y)$.
[guided]
Suppose first that $X$ and $Y$ are isomorphic in $\mathcal C$. This means that there are morphisms
\begin{align*}
f:X\to Y,
\qquad
g:Y\to X
\end{align*}
which are inverse to each other, so
\begin{align*}
g\circ f=\operatorname{id}_X,
\qquad
f\circ g=\operatorname{id}_Y.
\end{align*}
The point of applying $F$ is that functors preserve exactly the structure needed to recognize an isomorphism: they preserve identity morphisms and composition. Applying $F$ to the first identity gives
\begin{align*}
F(g)\circ F(f)
=
F(g\circ f)
=
F(\operatorname{id}_X)
=
\operatorname{id}_{F(X)}.
\end{align*}
Applying $F$ to the second identity gives
\begin{align*}
F(f)\circ F(g)
=
F(f\circ g)
=
F(\operatorname{id}_Y)
=
\operatorname{id}_{F(Y)}.
\end{align*}
Thus $F(f)$ has two-sided inverse $F(g)$, so $F(f):F(X)\to F(Y)$ is an isomorphism in $\mathcal D$. Hence $F(X)\cong F(Y)$.
[/guided]
[/step]
[step:Lift an isomorphism backward using fullness and prove the lifted morphisms are inverse using faithfulness]
Assume $F(X)\cong F(Y)$ in $\mathcal D$. Choose morphisms
\begin{align*}
u:F(X)\to F(Y),
\qquad
v:F(Y)\to F(X)
\end{align*}
such that
\begin{align*}
v\circ u=\operatorname{id}_{F(X)},
\qquad
u\circ v=\operatorname{id}_{F(Y)}.
\end{align*}
Since $F$ is full, the maps
\begin{align*}
F_{X,Y}:\operatorname{Hom}_{\mathcal C}(X,Y)&\to \operatorname{Hom}_{\mathcal D}(F(X),F(Y)),\\
h&\mapsto F(h),
\end{align*}
and
\begin{align*}
F_{Y,X}:\operatorname{Hom}_{\mathcal C}(Y,X)&\to \operatorname{Hom}_{\mathcal D}(F(Y),F(X)),\\
k&\mapsto F(k),
\end{align*}
are surjective. Hence there exist morphisms
\begin{align*}
f:X\to Y,
\qquad
g:Y\to X
\end{align*}
such that
\begin{align*}
F(f)=u,
\qquad
F(g)=v.
\end{align*}
We prove that $f$ and $g$ are inverse. By functoriality,
\begin{align*}
F(g\circ f)
=
F(g)\circ F(f)
=
v\circ u
=
\operatorname{id}_{F(X)}
=
F(\operatorname{id}_X).
\end{align*}
Since $F$ is faithful, the map $F_{X,X}:\operatorname{Hom}_{\mathcal C}(X,X)\to \operatorname{Hom}_{\mathcal D}(F(X),F(X))$ is injective. Therefore
\begin{align*}
g\circ f=\operatorname{id}_X.
\end{align*}
Similarly,
\begin{align*}
F(f\circ g)
=
F(f)\circ F(g)
=
u\circ v
=
\operatorname{id}_{F(Y)}
=
F(\operatorname{id}_Y),
\end{align*}
and faithfulness of $F_{Y,Y}$ gives
\begin{align*}
f\circ g=\operatorname{id}_Y.
\end{align*}
Thus $f:X\to Y$ is an isomorphism in $\mathcal C$, so $X\cong Y$.
[guided]
Now assume that $F(X)$ and $F(Y)$ are isomorphic in $\mathcal D$. Choose inverse morphisms
\begin{align*}
u:F(X)\to F(Y),
\qquad
v:F(Y)\to F(X)
\end{align*}
with
\begin{align*}
v\circ u=\operatorname{id}_{F(X)},
\qquad
u\circ v=\operatorname{id}_{F(Y)}.
\end{align*}
We want to recover an isomorphism between $X$ and $Y$ inside $\mathcal C$. The hypothesis that $F$ is an equivalence gives, in particular, that $F$ is full and faithful. Fullness means that every morphism between objects of the form $F(A)$ and $F(B)$ comes from a morphism $A\to B$ in $\mathcal C$. Applying fullness to
\begin{align*}
u\in \operatorname{Hom}_{\mathcal D}(F(X),F(Y))
\end{align*}
gives a morphism $f:X\to Y$ such that $F(f)=u$. Applying fullness to
\begin{align*}
v\in \operatorname{Hom}_{\mathcal D}(F(Y),F(X))
\end{align*}
gives a morphism $g:Y\to X$ such that $F(g)=v$.
It remains to prove that these lifted morphisms are inverse. We cannot merely assert this from fullness; fullness gives existence of lifts, not equations between lifts. The equations come from faithfulness. First compute:
\begin{align*}
F(g\circ f)
=
F(g)\circ F(f)
=
v\circ u
=
\operatorname{id}_{F(X)}
=
F(\operatorname{id}_X).
\end{align*}
The two morphisms $g\circ f$ and $\operatorname{id}_X$ both lie in $\operatorname{Hom}_{\mathcal C}(X,X)$. Since $F$ is faithful, the map
\begin{align*}
F_{X,X}:\operatorname{Hom}_{\mathcal C}(X,X)\to \operatorname{Hom}_{\mathcal D}(F(X),F(X))
\end{align*}
is injective. Therefore equality after applying $F$ implies equality before applying $F$:
\begin{align*}
g\circ f=\operatorname{id}_X.
\end{align*}
The same argument applies to the other composite:
\begin{align*}
F(f\circ g)
=
F(f)\circ F(g)
=
u\circ v
=
\operatorname{id}_{F(Y)}
=
F(\operatorname{id}_Y).
\end{align*}
Faithfulness of
\begin{align*}
F_{Y,Y}:\operatorname{Hom}_{\mathcal C}(Y,Y)\to \operatorname{Hom}_{\mathcal D}(F(Y),F(Y))
\end{align*}
then gives
\begin{align*}
f\circ g=\operatorname{id}_Y.
\end{align*}
Thus $f$ and $g$ are inverse isomorphisms, so $X\cong Y$ in $\mathcal C$.
[/guided]
[/step]
[step:Restrict the endomorphism bijection to obtain the automorphism group isomorphism]
Fix an object $X\in\mathcal C$. Since $F$ is full and faithful, the map
\begin{align*}
F_{X,X}:\operatorname{Hom}_{\mathcal C}(X,X)&\to \operatorname{Hom}_{\mathcal D}(F(X),F(X)),\\
h&\mapsto F(h),
\end{align*}
is a bijection. By the first part of the proof, a morphism $h:X\to X$ is an isomorphism in $\mathcal C$ if and only if $F(h):F(X)\to F(X)$ is an isomorphism in $\mathcal D$. Therefore $F_{X,X}$ restricts to a bijection
\begin{align*}
\Phi_X:\operatorname{Aut}_{\mathcal C}(X)&\to \operatorname{Aut}_{\mathcal D}(F(X)),\\
f&\mapsto F(f).
\end{align*}
It remains to check that $\Phi_X$ is a group homomorphism. The group operation on $\operatorname{Aut}_{\mathcal C}(X)$ is composition in $\mathcal C$, and the group operation on $\operatorname{Aut}_{\mathcal D}(F(X))$ is composition in $\mathcal D$. For any $f,g\in\operatorname{Aut}_{\mathcal C}(X)$,
\begin{align*}
\Phi_X(g\circ f)
=
F(g\circ f)
=
F(g)\circ F(f)
=
\Phi_X(g)\circ \Phi_X(f).
\end{align*}
Also
\begin{align*}
\Phi_X(\operatorname{id}_X)
=
F(\operatorname{id}_X)
=
\operatorname{id}_{F(X)}.
\end{align*}
Hence $\Phi_X$ is a bijective group homomorphism, and therefore an isomorphism of groups.
[guided]
Fix an object $X\in\mathcal C$. The endomorphisms of $X$ form the set
\begin{align*}
\operatorname{Hom}_{\mathcal C}(X,X),
\end{align*}
and the automorphism group $\operatorname{Aut}_{\mathcal C}(X)$ is the subset consisting of the invertible endomorphisms. Since $F$ is full and faithful, the induced map on endomorphism sets
\begin{align*}
F_{X,X}:\operatorname{Hom}_{\mathcal C}(X,X)&\to \operatorname{Hom}_{\mathcal D}(F(X),F(X)),\\
h&\mapsto F(h)
\end{align*}
is a bijection.
We now restrict this bijection to the invertible elements. The first part of the proof tells us exactly that invertibility is preserved and reflected by $F$: an endomorphism $h:X\to X$ is an isomorphism in $\mathcal C$ if and only if $F(h):F(X)\to F(X)$ is an isomorphism in $\mathcal D$. Therefore the bijection $F_{X,X}$ restricts to a bijection
\begin{align*}
\Phi_X:\operatorname{Aut}_{\mathcal C}(X)&\to \operatorname{Aut}_{\mathcal D}(F(X)),\\
f&\mapsto F(f).
\end{align*}
Finally, we verify that this bijection respects the group operations. The operation in each automorphism group is composition. For any automorphisms $f,g\in\operatorname{Aut}_{\mathcal C}(X)$, functoriality gives
\begin{align*}
\Phi_X(g\circ f)
=
F(g\circ f)
=
F(g)\circ F(f)
=
\Phi_X(g)\circ \Phi_X(f).
\end{align*}
Thus $\Phi_X$ preserves multiplication. It also preserves the identity element because
\begin{align*}
\Phi_X(\operatorname{id}_X)
=
F(\operatorname{id}_X)
=
\operatorname{id}_{F(X)}.
\end{align*}
So $\Phi_X$ is a bijective group homomorphism. Therefore $\Phi_X$ is an isomorphism of groups.
[/guided]
[/step]
