If $\theta$ is an involution of a complex linear algebraic group $G$ and if $K=G^\theta$ is its fixed-point set, then $K/K^0$ will always have exponent 2. This follows from a generalized "Cartan decomposition". The argument goes as follows. Let $\mathfrak p$ denote the (-1)-eigenspace of $d\theta$ on $\mathfrak g$. Then, morally at least, we ought to be able to express everything in $K$ as $k=k^0\exp X$, with $k_0 \in K^0$ and $X \in \mathfrak p$. But then $k^{-1} = (k_0\exp X)^{-1} = \exp(-X) (k_0)^{-1} = k' \exp(-X)$, for some $k' \in K^0$ (as $K^0$ is normal in $K$). On the other hand, $\exp(-X) = \exp(d\theta(X)) = \theta(\exp X) = \exp X$. Consequently, $k = k^{-1}$ in $K/K^0$, as desired. I've glossed over some details, which you can find in Loos, Symmetric Spaces, vol. 1, Benjamin (1969).

This follows from Theorem 8.1 in Steinberg, Endomorphisms of linear algebraic groups, Mem. Amer. Math. Soc. 80 (1968).

If $\theta$ is an involution of a complex linear algebraic group $G$ and if $K=G^\theta$ is its fixed-point set, then $K/K^0$ will always have exponent 2. This follows from a generalized "Cartan decomposition". The argument goes as follows. Let $\mathfrak p$ denote the (-1)-eigenspace of $d\theta$ on $\mathfrak g$. Then, morally at least, we ought to be able to express everything in $K$ as $k=k_0\exp X$, with $k_0 \in K^0$ and $X \in \mathfrak p$. But then $k^{-1} = (k_0\exp X)^{-1} = \exp(-X) k_0^{-1} = k' \exp(-X)$, for some $k' \in K^0$ (as $K^0$ is normal in $K$). On the other hand, $\exp(-X) = \exp(d\theta(X)) = \theta(\exp X) = \exp X$. Consequently, $k = k^{-1}$ in $K/K^0$, as desired. I've glossed over some details, which you can find in Loos, Symmetric Spaces, vol. 1, Benjamin (1969).

This follows from Theorem 8.1 in Steinberg, Endomorphisms of linear algebraic groups, Mem. Amer. Math. Soc. 80 (1968). An older reference---for compact Lie groups, at least---is Theorem 3.4 in Borel, Sous-groupes commutatifs et torsion des groupes de Lie compacts connexes, Tôhoku Math. J. 13 (1961), 216–240.