[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 $F(X)\cong F(Y)$.[/step]

[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: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 $X\cong Y$.[/step]

[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: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.[/step]

[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]