I posted the following question on MSE but I was advised to ask the question here.

Let $\mathfrak{g}$ be a lie algebra. Consider the anti-automomorphism

$$\phi: \mathfrak{g} \rightarrow \mathfrak{g} $$ such that $$\phi(X) = X^T, \quad \forall \quad X \in \mathfrak{g}$$ where $X^T$ is the usual transpose of $X$.

Observe that $$\phi([X,Y]) = \phi(XY-YX) = (XY-YX)^T = (XY)^T-(YX)^T.$$ Thus, we have that $$\phi([X,Y)] = Y^TX^T- X^TY^T = [\phi(Y), \phi(X)].$$

Note that we can extend this anti-automorphism to the universal enveloping algebra $\mathfrak{Ug}$ such that $\phi : \mathfrak{Ug} \rightarrow \mathfrak{Ug}$ still denoted by $\phi.$

**My question:**
Show that $$\phi(a)= a \quad$$ for all $a \in Z\mathcal{g} $ where $Z\mathcal{g}$ denotes the centre of $\mathfrak{Ug}.$

I thought about using the Harish-Chandra isomorphim but I dont seem to see how this helps.