Isometries $S$ on a finite-dimensional inner product space $V$ (over field $\mathbb{R}$ or $\mathbb{C}$) may or may not have a square root. I am expecting an equivalent condition such that $S$ has one.
SOME CONCEPTS
An isometry $S$ on an inner product space $V$ is a linear operator such that $||Sv||=||v||$ for all $v\in V$, where $||v||=\sqrt{\left<v,v\right>}$ denote the norm of $V$ induced by inner product.
A square root $R$ of a linear operator $T\in \operatorname{End}(V)$ is a linear operator on $V$ such that $R^2=T$.
MY PROGRESS
When field $\mathbb{F}=\mathbb{C}$, isometry $S\in \operatorname{End}(V)$ trivially has a square root, which result from the Spectrum Theorem.
When $\mathbb{F}=\mathbb{R}$, things become complex, as only real numbers are allowed, and an example of isometry without a square root is simple — just let $\dim V=1$ and $S=-I$.
$S\in \operatorname{End}(V)$ has a 'blocked diagonal matrix' $M$ with respect to some basis in $V$:
$$
M=
\begin{pmatrix}
B_1& & \\
& \ddots&\\
&&B_n
\end{pmatrix}
$$
where $B_1,\dots,B_n$ are:
a. $(1)$;
b. $(-1)$;
c. $2\times 2$ matrix
\begin{pmatrix}
\cos \theta &-\sin \theta\\
\sin \theta &\cos \theta
\end{pmatrix}
where $\theta\in (0,\pi)$.
The block c. is the rotation matrix (rotate $\theta$ counterclockwise) on $\mathbb{R}^2$, so its square root is the matrix rotating $\theta/2$ counterclockwise. For a. and b., if $1$ and $-1$ all appears even times, then just pair them (two $1$s or two $-1$s as a pair, not $1$ and $-1$ as a pair) and they also become rotation matrices (rotate $0$ and $\pi/2$) and their square roots are trivial. In this case, put all 'small' square roots together to form a new matrix and we find the 'big' square root of $S$.
If $1$ appears odd times but $-1$ appears even times, then leave one $1$ alone. Find 'small' square roots as above; as $1$'s square root is $1$ itself, the square root of $S$ is also easy to construct.
I have not made any progress when $-1$ appears odd times, because the method above fails in this case. I believe $S$ has no square roots now.
Best Answer
You've essentially done all the hard work already. To deal with the case you have left, compute the determinant.
The determinant of a block diagonal matrix is the products of the determinants, so you can find $\det(T)$ pretty easily from the form you've already given for $T$. Also notice that if $R^2 = T$ then $\det(R^2) = \det(R)^2 = \det(T)$, and in the real case $\det(R)$ is real, so $\det(T) \geq 0$.