[proofplan] We prove the two implications separately. In one direction, an isomorphism in $\mathcal{C}$ is sent by functoriality to an isomorphism in $\mathcal{D}$. In the reverse direction, since an equivalence of categories is full and faithful, an isomorphism $F(X)\to F(Y)$ and its inverse lift to morphisms $X\to Y$ and $Y\to X$, and faithfulness forces their composites to be the identity morphisms. [/proofplan] [step:Send an isomorphism in $\mathcal{C}$ to an isomorphism in $\mathcal{D}$] Assume $X\cong Y$ in $\mathcal{C}$. Choose an isomorphism $f:X\to Y$ in $\mathcal{C}$, and let $g:Y\to X$ denote its inverse, so that \begin{align*} g\circ f=\operatorname{id}_X, \qquad f\circ g=\operatorname{id}_Y. \end{align*} Applying the functor $F:\mathcal{C}\to\mathcal{D}$ and using preservation of identities and composition, we obtain \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*} Thus $F(f):F(X)\to F(Y)$ is an isomorphism in $\mathcal{D}$ with inverse $F(g):F(Y)\to F(X)$. Hence $F(X)\cong F(Y)$ in $\mathcal{D}$. [/step] [step:Lift an isomorphism in $\mathcal{D}$ using fullness] Assume $F(X)\cong F(Y)$ in $\mathcal{D}$. Choose an isomorphism $\alpha:F(X)\to F(Y)$ in $\mathcal{D}$, and let $\beta:F(Y)\to F(X)$ denote its inverse, so that \begin{align*} \beta\circ \alpha=\operatorname{id}_{F(X)}, \qquad \alpha\circ \beta=\operatorname{id}_{F(Y)}. \end{align*} Since $F$ is an equivalence of categories, it is full. Therefore the map \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*} is surjective, and the map \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*} is surjective. Hence there exist morphisms $f:X\to Y$ and $g:Y\to X$ in $\mathcal{C}$ such that \begin{align*} F(f)=\alpha, \qquad F(g)=\beta. \end{align*} [guided] We now use the part of equivalence called fullness. The isomorphism $\alpha:F(X)\to F(Y)$ lives in $\mathcal{D}$, but to prove that $X$ and $Y$ are isomorphic we need morphisms in $\mathcal{C}$. Fullness says exactly that every morphism between $F(X)$ and $F(Y)$ comes from a morphism between $X$ and $Y$. Since $F$ is full, the function \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*} is surjective. Applying this to $\alpha\in \operatorname{Hom}_{\mathcal{D}}(F(X),F(Y))$, we obtain a morphism $f:X\to Y$ with $F(f)=\alpha$. The inverse $\beta:F(Y)\to F(X)$ must also be lifted. Fullness applied to the pair $(Y,X)$ says that \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*} is surjective. Therefore there exists a morphism $g:Y\to X$ with $F(g)=\beta$. [/guided] [/step] [step:Use faithfulness to prove the lifted morphisms are inverse] It remains to show that $f$ and $g$ are inverse morphisms in $\mathcal{C}$. Using functoriality and the identities satisfied by $\alpha$ and $\beta$, we compute \begin{align*} F(g\circ f) &=F(g)\circ F(f)\\ &=\beta\circ \alpha\\ &=\operatorname{id}_{F(X)}\\ &=F(\operatorname{id}_X). \end{align*} 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))\\ h&\mapsto F(h) \end{align*} is injective. Thus $g\circ f=\operatorname{id}_X$. Similarly, \begin{align*} F(f\circ g) &=F(f)\circ F(g)\\ &=\alpha\circ \beta\\ &=\operatorname{id}_{F(Y)}\\ &=F(\operatorname{id}_Y). \end{align*} Faithfulness of $F$ on $\operatorname{Hom}_{\mathcal{C}}(Y,Y)$ gives $f\circ g=\operatorname{id}_Y$. Therefore $f:X\to Y$ is an isomorphism in $\mathcal{C}$ with inverse $g:Y\to X$, so $X\cong Y$. Combining the two implications proves that \begin{align*} X\cong Y \quad\Longleftrightarrow\quad F(X)\cong F(Y). \end{align*} [/step]