is this an orthogonal matrix? $T$ is a $4\times 4$ real matrix, and obeys $$T^\dagger \left(
\begin{array}{cccc}
 a & 0 & 0 & 0 \\
  0 & a & 0 & 0 \\
  0 & 0 & a^2 & 0 \\
  0 & 0 & 0 & a^2
\end{array}
\right) T = \left(
\begin{array}{cccc}
 a & 0 & 0 & 0 \\
  0 & a & 0 & 0 \\
  0 & 0 & a^2 & 0 \\
  0 & 0 & 0 & a^2
\end{array}
\right) $$ is $T$ an orthogonal matrix? If not, what ore do I need to ensure $T$ is orthogonal?
Other properties of $T$; $T^N=\mathbb{I}$ so $\det T=\pm 1$, $\alpha>1$.
 A: The matrix $T$ does not need to be orthogonal unless $a=1$. Something else can be though if $a>0$. You can write $T^\dagger A T=A$. If $A$ is positive definite (if $a>0$), it has a unique positive definite square root $A^{1/2}$ (in this case, a diagonal matrix with square roots of the corresponding diagonal entries on the diagonal). Multiplying by $A^{-1/2}$ from left and right then gives 
$$I=(A^{-1/2}T^\dagger A^{1/2})(A^{1/2}TA^{-1/2})=(A^{1/2}TA^{-1/2})^\dagger(A^{1/2}TA^{-1/2}),$$ which means that $\tilde T:=A^{1/2}TA^{-1/2}$ is orthogonal.

On the question "when is $T$ orthogonal":
Partition $T$ as
$$
T=\pmatrix{P&Q\\R&S},
$$
where the four blocks are square matrices ($2\times 2$). Then $T^\dagger AT=A$ gives (assuming $\alpha\neq 0$)
$$
P^\dagger P+\alpha R^\dagger R=I, \quad Q^\dagger Q+\alpha S^\dagger S=\alpha I, \quad
P^\dagger Q+\alpha R^\dagger S=0.
$$
Note that $T^\dagger T$ is equivalent to
$$
P^\dagger P+R^\dagger R=I, \quad Q^\dagger Q+S^\dagger S= I, \quad
P^\dagger Q+R^\dagger S=0.
$$
We want both to be true. Looking on the first two equations from both sets, we get
$$
(\alpha-1)R^\dagger R=0, \quad (\alpha-1)Q^\dagger Q=0.
$$
So either $\alpha=1$ or $R$ and $Q$ are zero matrices (or both). 
The case $\alpha=1$ is probably uninteresting, so let $\alpha\neq 1$ (and nonzero). Hence $R=Q=0$. The conditions above give that
$$
P^\dagger P=I, \quad S^\dagger S=I.
$$
So $T$ is block diagonal with orthogonal diagonal blocks.
On the other hand, if $T^\dagger A T=A$ holds for some $a\not\in\{0,1\}$ and $T$ is not block diagonal with orthogonal diagonal blocks, then $T$ is not orthogonal.
