What are the symmetric elements of $SU(3)$? Is there any simple way to know what are the matrices $M\in\operatorname{SU}(3)$ s.t $M=M^t$? For exemple, for $\operatorname{SU}(2)$ it is easy to verify that the symmetric elements are either diagonal or
$$\left(\begin{array}{cc} 
0 & i \\ 
i & 0
\end{array}\right).$$

I assume you are asking about the triplet representation, judging from your doublet paradigm for SU(2), which should not be "either-or", but rather
$$
i~ \begin{pmatrix}
                \sin\phi &  \cos\phi   \\
         \cos\phi & - \sin\phi
      \end{pmatrix} .
$$
I gather you are looking for "seat-of-the-pants" easy, so your first look should be at the elements generated by Gell-Mann's symmetric generators, {$\lambda_3, \lambda_8, \lambda_1, \lambda_4, \lambda_6$}, given the generic formula for the triplet rep group element,
$$
\begin{align}
  \exp(i\theta H) ={}
          &\left[-\frac{1}{3} I\sin\left(\varphi + \frac{2\pi}{3}\right) \sin\left(\varphi - \frac{2\pi}{3}\right) - \frac{1}{2\sqrt{3}}~H\sin(\varphi) - \frac{1}{4}~H^2\right]
           \frac{\exp\left(\frac{2}{\sqrt{3}}~i\theta\sin(\varphi)\right)}
                {\cos\left(\varphi + \frac{2\pi}{3}\right) \cos\left(\varphi - \frac{2\pi}{3}\right)} \\[6pt]
    & {} + \left[-\frac{1}{3}~I\sin(\varphi) \sin\left(\varphi - \frac{2\pi}{3}\right) - \frac{1}{2\sqrt{3}}~H\sin\left(\varphi + \frac{2\pi}{3}\right) - \frac{1}{4}~H^{2}\right]
           \frac{\exp\left(\frac{2}{\sqrt{3}}~i\theta \sin\left(\varphi + \frac{2\pi}{3}\right)\right)}
                {\cos(\varphi) \cos\left(\varphi - \frac{2\pi}{3}\right)} \\[6pt]
    & {}  + \left[-\frac{1}{3}~I\sin(\varphi) \sin\left(\varphi + \frac{2\pi}{3}\right) - \frac{1}{2\sqrt{3}}~H \sin\left(\varphi - \frac{2\pi}{3}\right) - \frac{1}{4}~H^2\right]
           \frac{\exp\left(\frac{2}{\sqrt{3}}~i\theta \sin\left(\varphi - \frac{2\pi}{3}\right)\right)}
                {\cos(\varphi)\cos\left(\varphi + \frac{2\pi}{3}\right)}
\end{align}
$$
generated by a traceless 3×3 Hermitian matrix  H, normalized as $\operatorname{tr}( H^2) =2$,  where
$$\varphi \equiv \frac{1}{3}\left[\arccos\left(\frac{3\sqrt{3}}{2}\det H\right) - \frac{\pi}{2}\right].$$
You notice that the linear combinations from this 5-generator set, squared, preserve the symmetry under transposition, so this subspace  generates elements in your set.
Exploring "coincidences" where special coefficients annihilate terms involving antisymmetric generators might be a formidable problem, however.
