= Solution
Take the definition of an <elementary topos> as a category with <finite limits>, <exponential objects> and a <subobject classifier> $\top:1\hookrightarrow\Omega$. Its <power object> is $PX=\Omega^X$, and $Pf:PY\to PX$ is precomposition with $f:X\to Y$. Transposing a predicate on $X\times Y$ in either variable gives
$$
\mathcal E(X,PY)\cong\mathcal E(Y,PX).
$$
Thus $P^{\mathrm{op}}:\mathcal E\to\mathcal E^{\mathrm{op}}$ is left adjoint to the <contravariant power-object functor> $P:\mathcal E^{\mathrm{op}}\to\mathcal E$.
Here are the remaining hypotheses for the <crude monadicity theorem>, obtained from those <topos> axioms. The unit of this <adjunction> is $\eta_X:X\to PPX$, internally $\eta_X(x)(S)=(x\in S)$. It is monic: if two such evaluations agree, evaluate on the singleton predicate $\{x\}$ classified by the diagonal to obtain $x=y$. If $Pf$ is invertible, naturality $PPf\,\eta_X=\eta_Yf$ therefore makes $f$ monic. The characteristic map of this mono, regarded as a global element of $PY$, pulls back along $f$ to the everywhere-true predicate on $X$. Since $Pf$ is monic, that characteristic map was already everywhere true on $Y$. The classifier <pullback> then says that $f$ is invertible. Hence $P$ is a <conservative functor>.
It remains to check preservation of <reflexive coequalizers> in the opposite category. Equivalently, let $f,g:B\rightrightarrows A$ be a <coreflexive pair>, with $r:A\to B$ satisfying $rf=rg=1_B$, and let $e:E\hookrightarrow B$ be their <equalizer>. Both $f$ and $g$ are monic. For any mono $i$, there is a direct-image map $\exists_i$ between <power objects>: a <subobject> is sent to its composite with $i$. This uses only classification of monos, not the prior existence of general images or colimits. We have $Pi\,\exists_i=1$.
For a predicate $W\hookrightarrow B$, consider its direct image along $f$. Its <pullbacks> along the two sections are
$$
Pf\,\exists_f(W)=W,\qquad
Pg\,\exists_f(W)=W\cap E=\exists_e Pe(W).
$$
To verify the second equality, $g(b)=f(w)$ with $w\in W$ implies $b=w$ after applying $r$, and therefore $f(b)=g(b)$; conversely $b\in W\cap E$ supplies that witness. This argument works for parameterized <subobjects> as well, so it is an equality of arrows between <power objects>.
Now if $h:PB\to Z$ satisfies $hPf=hPg$, then $h=h\exists_e Pe$. Hence $h\exists_e:PE\to Z$ is its unique factorization through $Pe$, uniqueness following from the section $\exists_e$. Therefore
$$
PA\mathrel{\substack{\xrightarrow{Pf}\\[-2pt]\xrightarrow[\ ]{Pg}}}PB\xrightarrow{Pe}PE
$$
is a <coequalizer>. <Finite limits> supply every required coreflexive <equalizer>. The right adjoint $P$ reflects isomorphisms and preserves the corresponding reflexive <coequalizers>, so \b[$P$ is monadic]. In particular $\mathcal E^{\mathrm{op}}$ is equivalent to the <Eilenberg-Moore category> of the double-power-object monad on $\mathcal E$.
Let $F:\mathcal E\to\mathcal F$ be a <logical functor>, and suppose $L\dashv F$. Preservation of exponentials and the classifier gives $F P_{\mathcal E}\cong P_{\mathcal F}F^{\mathrm{op}}$, compatibly with the units, counits and resulting monads. Thus $F^{\mathrm{op}}$ is the algebra <functor> lifting the base <functor> $F$ through the two monadic power-object <functors>. The <adjoint lifting theorem for monad algebra functors> applies to the base <adjunction> $L\dashv F$: its required <coequalizers> exist in $\mathcal E^{\mathrm{op}}$, because $\mathcal E$ has finite <equalizers>. It supplies a left adjoint $K\dashv F^{\mathrm{op}}$. Taking opposites gives \b[$F\dashv K^{\mathrm{op}}$], the required right adjoint to $F$.
Preserving either of the two logical structures by itself is insufficient. For the exponential example take \b[$F:\mathbf{Set}\to\mathbf{Set}$ constant at $1$]. The constant-empty <functor> is its left adjoint, since both relevant hom-sets are singletons. Its canonical exponential comparison is $1\cong1^1$, so it preserves exponentials. It has no right adjoint: a <functor> with a right adjoint would preserve the initial object, whereas $F(\varnothing)=1$.
For the classifier example take \b[$G:(\mathbb Z/2)\text{-}\mathbf{Set}\to\mathbf{Set}$ to be fixed points]. The trivial-action <functor> is left adjoint to $G$. In a group-action <topos> the classifier is the trivial-action two-element set, since invariant subsets have ordinary equivariant characteristic maps. Therefore $G$ preserves the classifier, its true arrow and the terminal object. But $G$ does not preserve the <coequalizer> of the identity and the nontrivial translation on the regular two-element group set: that <coequalizer> is $1$, while the fixed-point sets of the domain and codomain of the parallel pair are empty. Their set-theoretic <coequalizer> is empty, not $G(1)=1$. Thus $G$ has no right adjoint.
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