Let $G$ be a real simple Lie group of Hermitian type; that is, $G/K$ carries a structure of a Hermitian symmetric space where $K$ is a maximal compact subgroup of $G$. Equivalently, the center $Z(\mathfrak{k}_0)$ of Lie algebra $\mathfrak{k}_0$ of $K$ has dimension 1. Take $\theta$ to be the Cartan involution of $G$, which defines the maximal compact subgroup $K$.
Let $\tau$ be an involutive automorphism of $G$, which commutes with $\theta$. Use the same letter $\tau$ to denote its differential, and then $\tau$ stabilizes $\mathfrak{k}_0$ and also the center $Z(\mathfrak{k}_0)=\mathbb{R}Z$. Because $\tau^2=1$, there are two possibilities: $\tau Z=Z$ or $\tau Z=-Z$. Recall that the symmetric pair $(G,G^\tau)$ or its Lie algebra $(\mathfrak{g}_0,\mathfrak{g}_0^\tau)$ is said to be of holomorphic type (respectively, anti-holomorphic type) if $\tau Z=Z$ (respectively, $\tau Z=-Z$).
QUESTION Does the definition of holomorphic type depend on the choice of maximal compact subgroup $K$ or Cartan involution $\theta$?
Concretely, if there is another Cartan involution $\theta'$ of $G$, which commutes with $\tau$, then $\theta'=g^{-1}\theta g$ for some $g\in G$ and $\tau(\mathrm{Ad}(g)\mathfrak{k}_0)=\mathrm{Ad}(g)\mathfrak{k}_0$. Thus, $\mathrm{Ad}(g)\mathfrak{k}_0=\tau(\mathrm{Ad}(g)\mathfrak{k}_0)=\mathrm{Ad}(\tau(g))(\tau\mathfrak{k}_0)=\mathrm{Ad}(\tau(g))\mathfrak{k}_0$, and it follows that $\mathrm{Ad}(g^{-1}\tau(g))$ stabilizes $\mathfrak{k}_0$. In particular, $\mathrm{Ad}(g^{-1}\tau(g))$ stabilizes $Z(\mathfrak{k}_0)=\mathbb{R}Z$.
However, does $\mathrm{Ad}(g^{-1}\tau(g))Z=Z$ hold so that $\tau(\mathrm{Ad}(g)Z)=\mathrm{Ad}(\tau(g))\tau(Z)=\mathrm{Ad}(\tau(g))Z=\mathrm{Ad}(g)Z$? Namely, is $\tau$ identity on the center $\mathbb{R}\mathrm{Ad}(g)Z$ of the Lie algebra $\mathrm{Ad}(g)\mathfrak{k}_0$ of the maximal compact subgroup defined by $\theta'$?
FURTHER I know that it does not depend on the choices of maximal compact subgroups because according to the classification of symmetric pairs of holomorphic type and symmetric pairs of anti-holomorphic type, there does not exist a symmetric pair which is of both holomorphic type and anti-holomorphic type. However, I cannot deduce it by logical deduction as above. I shall be grateful if some expert here may give me any hint.
