Show that if $T_P X=P^{-1}XP$ for all $X$, then $(T_P)^*=T_{P^*}$ 
Let $V=M^{C}_{n \times n}$ with the standard inner product.
Suppose $T_{P}$ is a linear transformation from $V$ to $V$ such as: $\forall X \in V,\ T_{P}X=P^{-1}XP$ for some invertible matrix $P$.
Show that the following holds: $(T_{P})^{*}=T_{P^{*}}$ (that is, the conjugate transpose of T equals to T with P conjugate transposed).

In practice I need to show that $(T^{*}_{P})X=(P^*)^{-1}XP^*$, since $(P^{-1})^{*}=(P^{*})^{-1}$.
The examples I've seen so far showed that to find the conjugate transpose of a linear transformation, one needs to find the transformation matrix with respect to some orthonormal basis, and then use the relation $[T^*]_{B}=[T]_{B}^*$. But I somehow doubt that that needs to be done here.

*

*Are there any properties of the conjugate transpose of a linear transformation that can help me here?

*The linear transformation itself reminds me of matrix similarity, does it have any meaning?

 A: There are a couple of things you can do, depending on how you defined the adjoint of a linear operator.


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*If you defined $(T_P)^*$ as the unique linear transformation with the property that for all $A$ and $B$, $\langle T_P(A),B\rangle = \langle A,(T_P)^*(B)\rangle$, where $\langle\cdot,\cdot\rangle$ is the inner product of the space (which in this case is given by $\langle X,Y\rangle = \mathrm{Trace}(Y^*X)$, the Frobenius inner product, with $Y^*$ being the conjugate transpose). Then you need to show that $T_{P^*}$ has the "right" property. That is, that if $A$ and $B$ are matrices, then
$$\langle T_P(A),\rangle B = \langle A , T_{P^*}(B)\rangle.$$
So compute:
$$\begin{align*}
\langle T_P(A),B\rangle &= \langle P^{-1}AP,B\rangle = \mathrm{Trace}(B^*(P^{-1}AP)).\\
\langle A,T_{P^*}(B)\rangle &= \langle A,(P*)^{-1}BP\rangle = \mathrm{Trace}((P^*)^{-1}BP^*)^*A).
\end{align*}$$
Now use properties of the conjugate transpose and the trace to see if they are equal. 

*If you defined $(T_P)^*$ in terms of the matrix relative to an orthonormal basis, then take the standard orthonormal basis of $M_{n\times n}$, which consists of the matrices $E_{ij}$ that have a $1$ in the $(ij)$th entry and $0$s elsewhere, and see what $P^{-1}E_{ij}P$ is. 
The fact that it reminds you of matrix similarity is just the consequence that matrix similarity is "really" defined in terms of automorphisms: conjugating by an invertible matrix is an automorphism of the $n\times n$ matrices, and similarity consists of the orbits of the action of the group of these automorphisms. 
