The realification map $\varphi:\mathbb{C}^{n\times n}\rightarrow\mathbb{R}^{2n\times 2n}$ given by
$$Z=A+iB\mapsto\begin{bmatrix} A & -B\\ B & A \end{bmatrix}$$
provides an embedding of $\mathsf{U}(n)$ in $\mathsf{SO}(2n)$. It can be shown that $\varphi \mathsf{U}(n)=\mathsf{SO}(2n)\cap\varphi \mathsf{GL}(n,\mathbb{C})$ (I found this result in a textbook).
I am wondering if the orthogonal projection
$$\pi_{\varphi U}:\mathbb{R}^{2n\times 2n}\rightarrow\mathsf{T}_{\varphi U}\varphi \mathsf{U}(n):X\mapsto\pi_{\varphi U}X$$
where $U\in\mathsf{U}(n)$, restricted to $X\in\varphi\mathsf{GL}(n,\mathbb{C})$, is equal to the orthogonal projection
$$\Pi_R:\mathbb{R}^{2n\times 2n}\rightarrow\mathsf{T}_R\mathsf{SO}(2n):X\mapsto (X-X^T)R,$$
restricted to $X\in\varphi\mathsf{GL}(n,\mathbb{C})$ (assuming $R=\varphi U$)?
Attempt: I can show that $\Pi_R$ maps $\varphi\mathsf{GL}(n,\mathbb{C})$ to $\mathsf{T}_{\varphi U}\varphi \mathsf{U}(n)$ (details omitted). Assume there is an $X\in\mathbb{R}^{2n\times 2n}$ such that $\pi_{\varphi U}X\neq\Pi_RX$, where $R=\varphi U$. If one of $\|\Pi_RX-X\|$ and $\|\pi_{\varphi U}X-X\|$ is strictly smaller than the other, then this yields a contradiction: If $\|\pi_{\varphi U}X-X\|$ is smallest then this contradictions $\Pi_R$ being the orthogonal projection on $\mathsf{T}_R\mathsf{SO}(2n)$ and vice versa. It follows that $\|\Pi_RX-X\|=\|\pi_{\varphi U}X-X\|$. Hence both $\pi_{\varphi U}$ and $\Pi_R$ satisfy the condition of being orthogonal projections on $\varphi \mathsf{U}(n)$, and by the uniqueness of orthogonal projections on convex sets (tangent spaces are convex), the projections must be equal.
PS This is a follow-up question to my previous unanswered question (this not a repost since the question has changed):