I don't know if this is already answered somewhere in MO. The dynkin involution of $SL_{2n}$ that is $\alpha_i \mapsto \alpha_{2ni}$ gives an outer automorphism of $SL_{2n}$ and then the maximal parabolic $P_n$ corresponding to the simple root $\alpha_n$ is invariant under this automorphism. So it induces an automorphism of $Gr(n,2n)=SL_{2n}/P_n$. Do you get the Lagrangian Grassmannian from here say as a fixed points of this automorphism or you need to twist it by an inner automorphism of $SL_{2n}$ to get the Lagrangian Grassmannian $LG(n,2n)$ ? How is this automorphism related to the involution $V \mapsto V^{\perp}$ on $Gr(n,2n)$ ? Of course here one needs to fix a symplectic form on $\mathbb C^{2n}$ in order to define the map.
I think the confusion comes from the term "gives an outer automorphism" which has to be made precise. The automorphism, say $\phi$, is only welldefined up to multipliciation by conjugation with an element of the torus. To make it unique one usually requires that a "pinning" $e_\alpha$ is fixed. For the $SL(n)$ this means that, on the Lie algebra level, the elementary matrix $E_{i,i+1}$ is mapped to $E_{2ni,2ni+1}$. It is easily verfied that $\phi(A)=J^{1}(A^t)^{1}J$ does the job where $J=\text{antidiag}(1,1,1,\ldots,1)$.
Observe that $J$ is swewsymmetric so the fixed point set of $\phi$ is a symplectic group. To get the symplectic form $\omega$ one wants $A\mapsto\phi(A)^{1}$ to be the adjoint $A^*$ with respect to $\omega$: $$ \omega(Ax,y)=\omega(x,\phi(A)^{1}y)\text{ or, equivalently, }\omega(Ax,\phi(A)y)=\omega(x,y). $$ Another easy calculation shows that $\omega(x,y)=x^tJy$ fits. Any other solution is a nonzero multiple.
At this stage, the perpendicular space $U^\perp$ with respect to $\omega$ is defined. It is intertwined with $\phi$ via the formual $$ (AU)^\perp=\phi(A)U^\perp. $$ Thus if you let $\phi$ act on $Gr(n,2n)$ as $\phi(U)=U^\perp$ then one gets $\phi(AU)=\phi(A)\phi(U)$ and then $LG(n,2n)$ is precisely the fixed point set of $\phi$.

1$\begingroup$ Thank you for your beautiful answer. Indeed, that was the confusion. Combinatorially, how does the fixed points of $\phi$ (in the 1st paragraph of your answer) on $SL_{2n}/P_n$ look like ? How is it embedded in $Gr(n,2n)$ ? $\endgroup$ – Mark May 18 '17 at 18:43