Let $\mathbb C$ be a Cartesian closed category and $(T, \eta, \mu)$ be a monad on $\mathbb C$.
The extension operator $\_^{\sharp} : Hom(X, TY) → Hom(TX, TY)$ is defined by: $$f^{\sharp} = \mu_{Y}\circ Tf$$ Is it possible to define a morphism in $\mathbb C$ of type $(T Y)^X \to (T Y)^{T X}$?
It's not always possible to define such a morphism.
Regard the 4-element Boolean algebra $\{ \emptyset, \{0\}, \{1\}, \{0,1\} \}$, ordered by subset inclusion, as a poset category $\mathbb{B}_4$. This category is Cartesian closed, with product given by intersection, and exponential $X^Y$ given by $(\{0,1\}\setminus X) \cup Y$.
Consider the endofunctor $T: \mathbb{B}_4 \rightarrow \mathbb{B}_4$ which sends the empty set to itself, and sends every nonempty set to $\{0,1\}$. We have $X \subseteq TX$ and $TTX \subseteq TX$ for any object $X$ of $\mathbb{B}_4$. The required naturality squares trivially commute, so $T$ induces a monad on $\mathbb{B}_4$.
Let $X=\{0\}$ and $Y=\emptyset$. Then $(TY)^X = \{1\} \cup \emptyset = \{1\}$, while $(TY)^{TX} = \emptyset \cup \emptyset = \emptyset$, and there is no morphism from $\{1\}$ to $\emptyset$ in $\mathbb{B}_4$.