Let $P \in \Re^{n \times n}$ be an invertible matrix. Define
$$V=\{A \in \Re^{n \times n} | P^{-1} A P \ \text{is a diagonal matrix}\}$$
We define the following mappings
$$L:V \mapsto \Re^{n \times n}: A \mapsto P^{-1} A P$$
and
$$K:\Re^{n \times n} \mapsto \Re^{n} : M \mapsto (M_{11}, M_{22},...,M_{nn})$$
Show that $K \circ L$ is an isomorphism.
To show that $K \circ L$ is an isomorphism we need to prove that K and L are linear and that $K \circ L$ has an inverse.
I know how to show linearity so let's focus on the later. We'll show that $K \circ L$ is injective and surjective.
There's a theorem stating that a linear map $L$ is injective iff $Ker(L)=\{0\}$.
Given $K \circ L:V \rightarrow R^n$, we want to show that $Ker(K \circ L)=\vec 0$. Let $A \in V$. By definition of the kernel, $(K \circ L)(A)=0$. Then we want to see for what values of $A$ does $P^{-1}AP=0$ hold.
$$A=P(P^{-1}AP)P^{-1}=0 \Rightarrow A=0$$
As $0=A \in V \Rightarrow Ker(L)=\{0\} \Rightarrow L \ \text{is injective}$
We're only left to show that $K \circ L$ is onto. To do so, we need to prove that $Im(K \circ L)=\Re^n$. Let the n-tuple $(m_{11}, m_{22}, ..., m_{nn}) \in \Re^n$, which represents the diagonal entries of a diagonal matrix $M$. Thus by definition of the image (let $M \in \Re^n | (K \circ L)(M)=M$ where $M \in V$), $(K \circ L)(M)=M$. Then we want to satisfy the equation $P^{-1}MP=M$
$$P^{-1}MP=M \Rightarrow MP=PM$$
Mmm this is interesting. So it seems that $Im(K \circ L)=\Re^n$ only if $P,M$ commute. Do you agree or I am missing something? Are we done or we must show something else?
EDIT Let's try TheSilverDoe's approach:
Show that the inverse of $K \circ L:V \rightarrow \Re^n$ exists (i.e. $F:\Re^n \rightarrow V$). By definition of two the product of two maps we have
$$(K \circ L) \circ F : \Re^n \rightarrow \Re^n=Id_{\Re^n}$$.
$$F \circ (K \circ L) : V \rightarrow V=Id_V$$
Show $F:\Re^n \rightarrow V$ is linear.
Let $a,b \in \Re^n$ and $\lambda_1, \lambda_2 \in \Re$. We note that $a,b$ are the diagonal entries of two diagonal matrices $A,B$. Thus we have
$$F(\lambda_1 A+\lambda_2 B) = P^{-1}(\lambda_1 A+\lambda_2 B)P = \lambda_1 P^{-1} A P + \lambda_2 P^{-1} B P = \lambda_1 F(A) + \lambda_2 F(B)$$
EDIT 2.0
Let's show $(K \circ L) \circ F(x)=x$, where $x=(a_1,a_2,...,a_n) \in \Re^n$. We'll go from right to left.
$$F(x)=P \text{Diag}(a_1,a_2,...,a_n) P^{-1}$$
$$L(F(x))=P^{-1}(P \text{Diag}(a_1,a_2,...,a_n) P^{-1})P=\text{Diag}(a_1,a_2,...,a_n)$$
$$K(L(F(x)))=(a_1,a_2,...,a_n)=x$$
QED.
Let's show $F \circ (K \circ L)(A) =A$, where $A \in V$. We'll go from right to left.
$$L(A)=P^{-1} A P:=M$$
$$K(L(A))=(M_{11},M_{22},...,M_{nn})$$
$$F(K(L(A)))=PMP^{-1}=P(P^{-1} A P)P^{-1}=A$$
QED.
In my opinion, the most straightforward way is the following. Define $F : \mathbb{R}^n \rightarrow V$ by $$F(a_1, ..., a_n)= P \ \mathrm{Diag}(a_1, ..., a_n) \ P^{-1}$$
Prove that $F$ is linear and well-defined (i.e. why its image is included in $V$ ?).
Then, check that $(K \circ L) \circ F = \mathrm{Id}_{\mathbb{R}^n}$ and $F \circ (K \circ L) = \mathrm{Id}_{V}$