The second one is definitely correct. You can define a relation by its pairs.
$$
R = \left\{(a,a) : a \in A \right\}
$$
Reading the first one, it looks like you're saying that $a$ is related to $a$ for every $a$, and there is no $b$ not equal to $a$ that is related to $a$. That might look more like this:
$$
\forall a \in A,\left( a \mathrel{R} a\right)
\wedge \neg \left(\exists b \in A,\ (a \neq b) \wedge (a \mathrel{R} b)\right)
$$
That second part can be simplified logically, though.
\begin{align*}
\neg \left(\exists b \in A,\ (a \neq b) \wedge (a \mathrel{R} b)\right)
&\equiv \forall b \in A, (a=b) \vee (a \not\mathrel{R} b) \\
&\equiv \forall b \in A, (a \mathrel{R} b) \implies (a = b)
\end{align*}
That last part is because $p \implies q$ is equivalent to $(\neg p) \vee q$.
The first part can be combined with it:
\begin{align*}
\forall a \in A,\ a \mathrel{R} a
\equiv \forall a \in A,\ \forall b \in A,\ \left(a = b \implies a \mathrel{R} b\right)
\end{align*}
So the two together say:
\begin{align*}
&\forall a \in A,\ \forall b \in A,\
\left((a = b) \implies (a \mathrel{R} b)\right)
\wedge
\left((a \mathrel{R} b) \implies (a = b)\right)
\\&\equiv\forall a \in A,\ \forall b \in A,\
\left((a = b) \iff (a \mathrel{R} b)\right)
\end{align*}