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The goal is to write down explicit expressions of inner / outer automorphism of SU($n$), for $ngeq 2$.

We know that SU(2) has an SO(3) ($supseteq mathbb{Z}_2$)-inner automorphism,

while SU(n) has a $mathbb{Z}_2$-outer automorphism. For simply connected simple Lie groups, the outer automorphisms come from the automorphisms of the Dynkin diagram. See also the discussion in MO.

• For SU(2), we can write the group element as
  $$ g_{text{SU(2)}} = expleft(thetasum_{k=1}^{3} i t_k frac{sigma_k}{2}right) $$
  where $(t_1,t_2,t_3)$ forms a unit vector [effectively pointing in some direction on a unit 2-sphere $S^2$], and $sigma_k$ are Pauli matrices:
  begin{align}
  sigma_1 &=
  begin{pmatrix}
  0&1\
  1&0
  end{pmatrix} \
  sigma_2 &=
  begin{pmatrix}
  0&-i\
  i&0
  end{pmatrix} \
  sigma_3 &=
  begin{pmatrix}
  1&0\
  0&-1
  end{pmatrix} ,.
  end{align}
  Notice that any group element on $SU(2)$ can be parametrized by some $theta$ and $(t_1,t_2,t_3)$. Also $theta$ has a periodicity $[0,4 pi)$.

The inner automorphism is given by,
$$
x g_{text{SU(2)}} x^{-1}=
expleft(thetasum_{k=1}^{3} (-i) t_k frac{sigma_k^T}{2}right)
expleft(thetasum_{k=1}^{3} (-i) t_k frac{sigma_k^*}{2}right)
=g_{text{SU(2)}}^*.
$$
where
$$x=e^{ifrac{pi }{2}sigma_2} = isigma_2= begin{pmatrix}
0&1\
-1&0
end{pmatrix} in text{SU(2)},$$

• For SU($n$), $n>2$,

Do we have a simple expression of $g_{text{SU(n)}}$?

(It looks like the answer given here in ME by Anon is negative. But the Refs here Ref 1, Ref 2, Ref 3 writing down suggestive expressions
$$ g_{text{SU(n)}} = expleft(thetasum_{k=1}^{n^2-1} i t_k frac{lambda_k}{2}right)??? $$

So the outer automorphism of SU(n) simply sends $g_{text{SU(n)}}$ to its complex conjugation
$$
g_{text{SU(n)}} to g_{text{SU(n)}}^*?
$$

What is the explicit $x$ such that, for $n=3,4,5, etc$?
$$
g_{text{SU(n)}} to g_{text{SU(n)}}^* = x g_{text{SU(n)}} x^{-1}?
$$

A complaint first about notation: I learned to use the notation $g^{ast}$ for the complex conjugate transpose; you seem to be using it for just complex conjugate. To avoid this issue, I'll write $overline{g}$ for the complex conjugate of the matrix $g$.

The outer automorphism of $SU(n)$ is, indeed, $g mapsto overline{g}$. By the defining property of unitary matrices, this is also $g mapsto (g^T)^{-1}$. If $H$ is a Hermitian matrix, then this automorphism sends $exp(iH)$ to $exp(-ioverline{H}) = exp(-i H^T)$. Of course, you can express this formula in terms of any basis for the Hermitian matrices you like.

Once $n$ is at least $3$, the matrices $g$ and $overline{g}$ are generically not conjugate. Let the eigenvalues of $g$ be $exp(i theta_1)$, $exp(i theta_2)$, .., $exp(i theta_n)$. Then the eigenvalues of $overline{g}$ will be $exp(-i theta_1)$, $exp(-i theta_2)$, .., $exp(-i theta_n)$. For generic $(theta_1, ldots, theta_n)$ with $sum theta_j=0$, the second list of eigenvalues will not be a permutation of the first, so $g$ and $overline{g}$ are not conjugate within $SU(n)$ (or even $GL_n$). Indeed, this is the easiest way to see that the automorphism is outer. So your request for a matrix $x$ with $overline{g} = x g x^{-1}$ doesn't make sense.