$F(x,y)$ : $x$ can fool $y$.
Domain : All People
There is exactly one person whom everybody can fool.
1. Using Uniqueness Quantifier ($\exists !$) :
$\exists ! x \forall y(F(y,x))$ : There is Exactly One Person $x$ such that All persons $y$ can fool $x$.
2. Using $\exists, \forall$ :
$\exists x \forall y(F(y,x) \wedge \forall z((z \neq x) \rightarrow \exists p(\sim F(p,z))))$
Interpretation : There exists some $x$ such that for all persons $y$, $y$ can fool $x$ and for all other persons $z$( $z \neq x$), there is some person $p$ who cannot fool $z$.
Note that : "There exists some $x$ such that for all persons $y$, $y$ can fool $x$" will make sure that there is at least one $x$ whom everyone can fool.
and "for all other persons $z$( $z \neq x$), there is some person $p$ who cannot fool $z$." will make sure that apart from $x$, there is No person whom everyone can fool.
Since we know that Every Implication statement is equivalent to its Contrapositive, the above formula can also be written as the following :
$\exists x \forall y(F(y,x) \wedge \forall z((\forall p F(p,z) \rightarrow (z = x)))$
3. Another idea to find expression for "Exactly One" is as following :
Exactly One = At least One - At least Two (In logic, Minus can be expressed by ($∧,∼$)
$\exists x \forall y(F(y,x)) $ $\wedge$ $(\sim \exists x \exists y \forall z ((x \neq y) \wedge F(z,x) \wedge F(z,y)))$
There is someone who can fool exactly one person besides himself or herself.
Assuming the Question wants to say that Someone($x$) can fool exactly one other person($y$) and $X$ can not fool himself/herself.
$\exists x \exists y((x \neq y) \wedge F(x,y) \wedge \forall z ((z \neq y) \rightarrow \sim F(x,z)))$
There is some $x$ and some $y$ such that $x \neq y$ and $x$ can fool $y$ and For all people $z$ other than $y$, $x$ cannot fool $z$.
Note 1 : "There is some $x$ and some $y$ such that $x \neq y$ and $x$ can fool $y$" will make sure that there is someone($x$) who can fool someone($y$) other than himself/herself($x \neq y$ ). and "and For all people $z$ other than $y$, $x$ cannot fool $z$. " will make sure that $x$ can not fool any other person $x$ other than $y$.
Note 2 : If some person $x$ can fool himself and exactly one other person $y$ then we can just slightly modify our formula :
$\exists x \exists y((x \neq y) \wedge F(x,y) \wedge F(x,x) \wedge \forall z ((z \neq y,x) \rightarrow \sim F(x,z)))$
Refer here for more details on "Exactly one" type question : https://gateoverflow.in/219473/kenneth-rosen-ch-1-ex-1-5-qn-52?show=219480#a219480
