I haven't really studied complex manifolds and I am at a bit of a loss in regards how to approach this problem:
Let $M$ be an $n$-dimensional complex manifold, and let $\phi:M\rightarrow M$ be an antiholomorphic involution ($\phi^2 = id$, and $\phi$ is defined by antiholomorphic functions in local charts on $M$) of $M$. Suppose that the fixed point set of $\phi$, $N := \{y\in M \: :\: \phi(y) =y\}$, is an $n$-dimensional (real) submanifold of $M$ (where we view $M$ as a $2n$-dimensional real smooth manifold).
I want to show that the complexified tangent space, $T_y(N)\otimes \mathbb{C}$, is canonically isomorphic to $T_yM$, in the sense that the isomorphism between the space commutes with maps induced by complex homeomorphisms that preserve $y$ and commute with $\phi$.
Thoughts:
Let $\{z_1,...,z_n\}$ be coordinates around $y$. Writing $z_i = x_i + iy_i$ we can write a basis for $T_yM$, as a vector space over $\mathbb{R}$, by $$\{ \frac{\partial}{\partial x^1},...,\frac{\partial}{\partial x^n},\frac{\partial}{\partial y^1},...,\frac{\partial}{\partial y^n} \}.$$
A basis for $T_y(N)\otimes \mathbb{C}$ over $\mathbb{R}$ is given by $$\{\frac{\partial}{\partial x^1}\otimes 1,...,\frac{\partial}{\partial x^n}\otimes 1,\frac{\partial}{\partial x^1}\otimes i,...,\frac{\partial}{\partial x^n}\otimes i\}.$$
We can then define the isomorphism by mapping $\frac{\partial}{\partial x^j}\otimes 1$ to $\frac{\partial}{\partial x^j}$ and mapping $\frac{\partial}{\partial x^j}\otimes i$ to $\frac{\partial}{\partial y^j}$.
Am I on the right track? I'm not sure how to show that such an isomorphism is canonical or that I understand given definition of canonical.
Hint: If you continue with what you've done until now, then, if you can choose a specific isomorphism such that $$\dfrac{\partial }{\partial x^j}\mapsto \dfrac{\partial }{\partial x^j}\otimes1\\ \dfrac{\partial }{\partial y^j}\mapsto \dfrac{\partial }{\partial x^j}\otimes (-i)\tag{1}$$ Then you are done. I.e., showing that $(1)$ is a bijective homomorphism is sufficient.