What is the infimal norm of a matrix in a conjugacy class? $\newcommand{\tr}{\operatorname{tr}}$Let $A$ be a real $d \times d$ matrix. What is 
$$ F(A):=\inf_{M \in \text{GL}(d, \mathbb{R})} |MAM^{-1}|^2?$$
Here $| \cdot |$ is the standard Euclidean (Frobenius) norm. Is the infimum always obtained? I would like to find a simple formula for $F(A)$ in terms of $A$.
Cauchy-Schwarts inequality implies that $$( \tr(A))^2 \le d |A|^2$$ with equality if and only if $A$ is a multiple of the identity matrix. 
Since the trace is an invariant of a conjugacy class, we deduce
$$ F(A) \ge \frac{\tr(A)^2}{d},$$ and this value is obtained if  and only if $A$ is a multiple of the identity matrix. 
Comment: Note that I am referring to conjugacy classes in $\mathrm{GL}(d, \mathbb{R})$, not in $\mathrm{GL}(d, \mathbb{C})$.

Not a proof, but an idea: 
Claim 1: It suffices to consider $A$ in Jordan canonical form (upper triangular)
Claim 2: $F(A)\leq \|A \|_2^2$ 
Proof: Consider the diagonal matrix $D_t=\mathrm{diag}(t,t^2,\dots,t^d)$ for $t>0$. We compute $D_t^{-1}AD_t$. The diagonal still has the eigenvalues, while the off-diagonal terms are multiplied by a positive power of $t$. Letting $t\rightarrow 0$ we have  $D_t^{-1}AD_t$ converging to a diagonal matrix with the eigenvalues on the diagonal, and the result follows. 
Note: $\| A\|_2^2=\frac{\mathrm{tr}(A)^2}{d}$ for multiples of the identity. My guess is that $F(A)= \|A \|_2^2$ and the infimum is not a min unless the Jordan form of $A$ is diagonal.
Note: I have assumed the inf is over $\mathrm{GL}(d, \mathbb{C})$. If instead you meant $\mathrm{GL}(d, \mathbb{R})$, then this is all probably useless.
