I will post a partial answer that may reduce the problem to something more familiar to others (that I am not familiar with).

The map $\mathfrak{t}^{\perp} \to T_{\lambda}\mathcal{O}_{\lambda}$, $X \mapsto ad_X(\lambda)$, endows $\mathfrak{t}^{\perp}$ with a linear symplectic form $$ \omega_0(X,Y) = \langle \lambda, [X,Y]\rangle.$$ Based on edit 2 above, we know there exists a $T$-equivariant map $\phi\colon \mathfrak{t}^{\perp} \to \mathfrak{t}^{\perp}$ (at least, defined and invertible near 0) such that $\phi(0) = 0$ and the composition $$\varphi(Y) := Ad_{e^{\phi(Y)}}\lambda$$ is a symplectomorphism. Let's use this knowledge to try to learn something more about $\phi$.

Let $\pi\colon K \to \mathcal{O}_{\lambda}$ be the map $k \mapsto Ad_k\lambda$. We know that

• $d(\pi)_e Y = ad_Y\lambda$
• $d(\pi)_k = d(Ad_k)_{\lambda} \circ d(\pi)_e \circ d(\mathcal{L}_{k^{-1}})_k$ (where $\mathcal{L}_k\colon K \to K$ is left multiplication).
• $d(\mathcal{L}_{k^{-1}})_k$ is simply the (left-invariant) Maurer-Cartan form $\theta$ evaluated at $k$
• Because it is a linear map, $d(Ad_k)_{\lambda} = Ad_k$. We also know that $Ad_kad_X \lambda = ad_{Ad_kX}Ad_k\lambda$.

Thus:
\begin{equation} \begin{split} d(\varphi)_Y & = d(\pi)_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\\ & = d(Ad_{e^{\phi(Y)}})_{\lambda} \circ d(\pi)_e \circ d(\mathcal{L}_{e^{-\phi(Y)}})_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\\ & = Ad_{e^{\phi(Y)}}\left(d(\pi)_e \circ \theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\right)\\ & = Ad_{e^{\phi(Y)}}\left(ad_{\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y}\lambda\right)\\ & = ad_{Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\right)}\left(Ad_{e^{\phi(Y)}}\lambda\right). \end{split} \end{equation}

The map $\varphi$ is a symplectomorphism if $\varphi^* \omega = \omega_0$. Explicitly, for $X,Z \in \mathfrak{t}^{\perp}$, $$(\varphi^*\omega)_Y(X,Z) = \omega_{\varphi(Y)}\left(d(\varphi)_Y X,d(\varphi)_Y Z\right)$$ which by the previous equation $$ = \langle Ad_{e^{\phi(Y)}}\lambda, [Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y X\right), Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y Z\right)]\rangle$$ which, since $\omega$ is $K$-invariant, $$ = \langle \lambda, [\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y X, \theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y Z]\rangle.$$ This equals $\omega_0(X,Z)$ (as written above) if it is true that $$\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y = \text{id}_{\mathfrak{k}}$$ There is a very similar equation (see formula (1.12) on page 16 of Sternberg's notes on Lie Algebras): $$ \tau(ad_Y) \circ \left(\theta_{\exp(Y)}\circ d(\exp)_Y\right) = \text{id}_{\mathfrak{k}}$$ where $\tau(w) = \frac{w}{1-e^{-w}}$. Thus we conclude that $\phi$ must solve the equation $$d\phi_Y = \tau(ad_{\phi(Y)}) = \text{id} + \frac{ad_{\phi(Y)}}{2}+\frac{ad_{\phi(Y)}^2}{12} + ...$$ I guess this is the point where I stop because I don't know how to solve this equation.

Here is a partial answer.

The map $\mathfrak{t}^{\perp} \to T_{\lambda}\mathcal{O}_{\lambda}$, $X \mapsto ad_X(\lambda)$, endows $\mathfrak{t}^{\perp}$ with a linear symplectic form $$ \omega_0(X,Y) = \langle \lambda, [X,Y]\rangle.$$ Based on edit 2 above, we know there exists a $T$-equivariant map $\phi\colon \mathfrak{t}^{\perp} \to \mathfrak{t}^{\perp}$ (at least, defined and invertible near 0) such that $\phi(0) = 0$ and the composition $$\varphi(Y) := Ad_{e^{\phi(Y)}}\lambda$$ is a symplectomorphism.

Example ($K = SU(2)$) Here we can compute using cylindrical coordinates on the coadjoint orbit and polar coordinates on $\mathfrak{t}^{\perp}$ to show that $$\phi\left(\left(\begin{array}{cc} 0 & \rho e^{i\theta} \\ -\rho e^{-i\theta} & 0 \end{array}\right)\right) = \left(\begin{array}{cc} 0 & \arcsin(\rho) e^{i\theta} \\ -\arcsin(\rho) e^{-i\theta} & 0 \end{array}\right).$$

Let $\pi\colon K \to \mathcal{O}_{\lambda}$ be the map $k \mapsto Ad_k\lambda$. We know that

• $d(\pi)_e Y = ad_Y\lambda$
• $d(\pi)_k = d(Ad_k)_{\lambda} \circ d(\pi)_e \circ d(\mathcal{L}_{k^{-1}})_k$ (where $\mathcal{L}_k\colon K \to K$ is left multiplication).
• $d(\mathcal{L}_{k^{-1}})_k$ is simply the (left-invariant) Maurer-Cartan form $\theta$ evaluated at $k$
• Because it is a linear map, $d(Ad_k)_{\lambda} = Ad_k$. We also know that $Ad_kad_X \lambda = ad_{Ad_kX}Ad_k\lambda$.

Thus:
\begin{equation} \begin{split} d(\varphi)_Y & = d(\pi)_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\\ & = d(Ad_{e^{\phi(Y)}})_{\lambda} \circ d(\pi)_e \circ d(\mathcal{L}_{e^{-\phi(Y)}})_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\\ & = Ad_{e^{\phi(Y)}}\left(d(\pi)_e \circ \theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\right)\\ & = Ad_{e^{\phi(Y)}}\left(ad_{\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y}\lambda\right)\\ & = ad_{Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\right)}\left(Ad_{e^{\phi(Y)}}\lambda\right). \end{split} \end{equation}

The map $\varphi$ is a symplectomorphism if $\varphi^* \omega = \omega_0$. Explicitly, for $X,Z \in \mathfrak{t}^{\perp}$, $$(\varphi^*\omega)_Y(X,Z) = \omega_{\varphi(Y)}\left(d(\varphi)_Y X,d(\varphi)_Y Z\right)$$ which by the previous equation $$ = \langle Ad_{e^{\phi(Y)}}\lambda, [Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y X\right), Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y Z\right)]\rangle$$ $$ = \langle \lambda, [\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y X, \theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y Z]\rangle.$$ This simplifies by substituting the formula for the derivative of the exponential map: $$\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)} = e^{-\phi(Y)}e^{\phi(Y)}\left(\frac{1-e^{-ad_{\phi(Y)}}}{ad_{\phi(Y)}}\right) = \frac{1-e^{-ad_{\phi(Y)}}}{ad_{\phi(Y)}}$$ to give $$(\varphi^*\omega_{\lambda})_Y(X,Z) = \bigg\langle \lambda, \frac{1-e^{-ad_{\phi(Y)}}}{ad_{\phi(Y)}}[ d\phi_Y X, d\phi_Y Z]\bigg\rangle.$$

Thus, our equation for $\phi$ is \begin{equation} \bigg\langle \lambda, \frac{1-e^{-ad_{\phi(Y)}}}{ad_{\phi(Y)}}[ d\phi_Y X, d\phi_Y Z]\bigg\rangle = \langle \lambda, [X,Z]\rangle. \end{equation}

The series expansion of the LHS will have some terms that disappear because they pair with $\lambda$ to zero, but I don't have a general formula for the ones that remain.

I will post a partial answer that may reduce the problem to something more familiar to others (that I am not familiar with).

The map $\mathfrak{t}^{\perp} \to T_{\lambda}\mathcal{O}_{\lambda}$, $X \mapsto ad_X(\lambda)$, endows $\mathfrak{t}^{\perp}$ with a linear symplectic form $$ \omega_0(X,Y) = \langle \lambda, [X,Y]\rangle.$$ Based on edit 2 above, we know there exists a $T$-equivariant map $\phi\colon \mathfrak{t}^{\perp} \to \mathfrak{t}^{\perp}$ (at least, defined and invertible near 0) such that $\phi(0) = 0$ and the composition $$\varphi(Y) := Ad_{e^{\phi(Y)}}\lambda$$ is a symplectomorphism. Let's use this knowledge to try to learn something more about $\phi$.

Let $\pi\colon K \to \mathcal{O}_{\lambda}$ be the map $k \mapsto Ad_k\lambda$. We know that

• $d(\pi)_e Y = ad_Y\lambda$
• $d(\pi)_k = d(Ad_k)_{\lambda} \circ d(\pi)_e \circ d(\mathcal{L}_{k^{-1}})_k$ (where $\mathcal{L}_k\colon K \to K$ is left multiplication).
• $d(\mathcal{L}_{k^{-1}})_k$ is simply the (left-invariant) Maurer-Cartan form $\theta$ evaluated at $k$
• Because it is a linear map, $d(Ad_k)_{\lambda} = Ad_k$. We also know that $Ad_kad_X \lambda = ad_{Ad_kX}Ad_k\lambda$.

Thus:
\begin{equation} \begin{split} d(\varphi)_Y & = d(\pi)_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\\ & = d(Ad_{e^{\phi(Y)}})_{\lambda} \circ d(\pi)_e \circ d(\mathcal{L}_{e^{-\phi(Y)}})_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\\ & = Ad_{e^{\phi(Y)}}\left(d(\pi)_e \circ \theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\right)\\ & = Ad_{e^{\phi(Y)}}\left(ad_{\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y}\lambda\right)\\ & = ad_{Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\right)}\left(Ad_{e^{\phi(Y)}}\lambda\right). \end{split} \end{equation}

The map $\varphi$ is a symplectomorphism if $\varphi^* \omega = \omega_0$. Explicitly, for $X,Z \in \mathfrak{t}^{\perp}$, $$(\varphi^*\omega)_Y(X,Z) = \omega_{\varphi(Y)}\left(d(\varphi)_Y X,d(\varphi)_Y Z\right)$$ which by the previous equation $$ = \langle Ad_{e^{\phi(Y)}}\lambda, [Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y X\right), Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y Z\right)]\rangle$$ which, since $\omega$ is $K$-invariant, $$ = \langle \lambda, [\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y X, \theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y Z]\rangle.$$ This equals $\omega_0(X,Z)$ (as written above) if it is true that $$\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y = \text{id}_{\mathfrak{k}}$$ There is a very similar equation (see formula (1.12) on page 16 of Sternberg's notes on Lie Algebras): $$ \tau(ad_Y) \circ \left(\theta_{\exp(Y)}\circ d(\exp)_Y\right) = \text{id}_{\mathfrak{k}}$$ where $\tau(w) = \frac{w}{1-e^{-w}}$. Thus we conclude that $\phi$ must solve the equation $$d\phi_Y = \tau(ad_{\phi(Y)}) = \text{id} + \frac{ad_{\phi(Y)}}{2}+\frac{ad_{\phi(Y)}}{12} + ...$$ I guess this is the point where I stop because I don't know how to solve this equation.

I will post a partial answer that may reduce the problem to something more familiar to others (that I am not familiar with).

The map $\mathfrak{t}^{\perp} \to T_{\lambda}\mathcal{O}_{\lambda}$, $X \mapsto ad_X(\lambda)$, endows $\mathfrak{t}^{\perp}$ with a linear symplectic form $$ \omega_0(X,Y) = \langle \lambda, [X,Y]\rangle.$$ Based on edit 2 above, we know there exists a $T$-equivariant map $\phi\colon \mathfrak{t}^{\perp} \to \mathfrak{t}^{\perp}$ (at least, defined and invertible near 0) such that $\phi(0) = 0$ and the composition $$\varphi(Y) := Ad_{e^{\phi(Y)}}\lambda$$ is a symplectomorphism. Let's use this knowledge to try to learn something more about $\phi$.

Let $\pi\colon K \to \mathcal{O}_{\lambda}$ be the map $k \mapsto Ad_k\lambda$. We know that

• $d(\pi)_e Y = ad_Y\lambda$
• $d(\pi)_k = d(Ad_k)_{\lambda} \circ d(\pi)_e \circ d(\mathcal{L}_{k^{-1}})_k$ (where $\mathcal{L}_k\colon K \to K$ is left multiplication).
• $d(\mathcal{L}_{k^{-1}})_k$ is simply the (left-invariant) Maurer-Cartan form $\theta$ evaluated at $k$
• Because it is a linear map, $d(Ad_k)_{\lambda} = Ad_k$. We also know that $Ad_kad_X \lambda = ad_{Ad_kX}Ad_k\lambda$.

Thus:
\begin{equation} \begin{split} d(\varphi)_Y & = d(\pi)_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\\ & = d(Ad_{e^{\phi(Y)}})_{\lambda} \circ d(\pi)_e \circ d(\mathcal{L}_{e^{-\phi(Y)}})_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\\ & = Ad_{e^{\phi(Y)}}\left(d(\pi)_e \circ \theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\right)\\ & = Ad_{e^{\phi(Y)}}\left(ad_{\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y}\lambda\right)\\ & = ad_{Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\right)}\left(Ad_{e^{\phi(Y)}}\lambda\right). \end{split} \end{equation}

The map $\varphi$ is a symplectomorphism if $\varphi^* \omega = \omega_0$. Explicitly, for $X,Z \in \mathfrak{t}^{\perp}$, $$(\varphi^*\omega)_Y(X,Z) = \omega_{\varphi(Y)}\left(d(\varphi)_Y X,d(\varphi)_Y Z\right)$$ which by the previous equation $$ = \langle Ad_{e^{\phi(Y)}}\lambda, [Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y X\right), Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y Z\right)]\rangle$$ which, since $\omega$ is $K$-invariant, $$ = \langle \lambda, [\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y X, \theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y Z]\rangle.$$ This equals $\omega_0(X,Z)$ (as written above) if it is true that $$\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y = \text{id}_{\mathfrak{k}}$$ There is a very similar equation (see formula (1.12) on page 16 of Sternberg's notes on Lie Algebras): $$ \tau(ad_Y) \circ \left(\theta_{\exp(Y)}\circ d(\exp)_Y\right) = \text{id}_{\mathfrak{k}}$$ where $\tau(w) = \frac{w}{1-e^{-w}}$. Thus we conclude that $\phi$ must solve the equation $$d\phi_Y = \tau(ad_{\phi(Y)}) = \text{id} + \frac{ad_{\phi(Y)}}{2}+\frac{ad_{\phi(Y)}^2}{12} + ...$$ I guess this is the point where I stop because I don't know how to solve this equation.