Does this specific SO(4) matrix have to be block-diagonal? So I have a specific real $4\times4$ matrix $\mathbf{P}$ given by 
\begin{align}
\mathbf{P}=
\begin{pmatrix}
p_{11} & -p_{21} & p_{13} &-p_{23}\\ p_{21} & p_{11} & p_{23} & p_{13}\\p_{31} & -p_{41}& p_{33} & -p_{43}\\p_{41} & p_{31} & p_{43} & p_{33}.
\end{pmatrix},
\end{align}
and I'm confident that if this matrix is in $SO(4)$ then it must be block-diagonal OR anti-block-diagonal i.e. if $\mathbf{P}\in SO(4)$ then either $p_{11}=p_{21}=p_{33}=p_{43}=0$ or $p_{13}=p_{23}=p_{31}=p_{41}=0$, but I can't seem to show this...  
I've tried picking one column and generating a set of orthonormal vectors to fill the matrix, but this just gives examples where it works and I'd like to show it generally.  Is this possible?  Am I just missing something trivial here?

Skip to the end for the big reveal, or read through this for the "how I got there" version. 
Let's rewrite that as 
\begin{align}
\mathbf{P}=
\begin{pmatrix}
a & -b & p &-q\\ 
b & a & q & p\\
c & -d& r & -s\\
d & c & s & r.
\end{pmatrix},
\end{align}
Orthogonality of the first and third and first and 4th columns tells us that 
\begin{align}
ap + bq + cr + ds &= 0\\
-aq + bp - cs + dr &= 0\\
\end{align}
Cross-multiply to get
\begin{align}
apq + bq^2 + crq + dsq &= 0 \\
-apq + bp^2 - cps + dpr &= 0\\
\end{align}
and sum to get
\begin{align}
b(p^2 + q^2) + crq - cps  + dsq + dpr &= 0 \\
\end{align}
Doing the same for columns 2 against 3 and 4 gives
\begin{align}
a(p^2 + q^2)  - drq + csq + dps + cpr &= 0 \\
\end{align}
Let's factor those to get
\begin{align}
b(p^2 + q^2) + c(rq - ps)  + d(sq + pr) &= 0 \\
a(p^2 + q^2) + c(sq + pr) - d(rq - ps)  &= 0 \\
\end{align}
Looking at the dot product between rows 2 and 3, we see that 
$$
rq - ps = ad - bc
$$
and similarly, for rows 2 and 4, we get
$$
qs + pr = - (ac + bd)
$$
so 
\begin{align}
b(p^2 + q^2) + c(ad-bc)  - d(ac + bd) &= 0 \\
a(p^2 + q^2) - c(ac + bd) - d(ad-bc)  &= 0 \\
\end{align}
which simplifies to 
\begin{align}
b(p^2 + q^2) - bc^2 - bd^2 &= 0 \\
a(p^2 + q^2) - ac^2 - ad^2  &= 0 \\
\end{align}
which become
\begin{align}
b(p^2 + q^2 - c^2 - d^2) &= 0 \\
a(p^2 + q^2 - c^2 - d^2) &= 0 \\
\end{align}
We conclude that either 
(1) $p^2 + q^2 = c^2 + d^2$ or
(2) $a = b = 0$.
In the second case, we have that the squared norm of the first row is $p^2 + q^2$, which must be $1$, and the squared norm of the first column is $c^2 + d^2$, which must also be $1$, hence $p^2 + q^2 = c^2 + d^2$. 
In other words, in all cases, $p^2 + q^2 = c^2 + d^2$. 
By looking at the squared norms of the first row and the 4th column, we find that 
$$
a^2 + b^2 = r^2 + s^2
$$
as well. 
But nothing else obvious seems to jump out...
...and so I began to wonder if it was actually true, and came up with this:
\begin{align}
\mathbf{P}=
\begin{pmatrix}
\frac{1}{2} & -\frac{1}{2} & 0 &-s\\ 
\frac{1}{2} & \frac{1}{2} & s & 0\\
\frac{1}{2} & -\frac{1}{2}& 0 & s\\
\frac{1}{2} & \frac{1}{2} & -s & 0
\end{pmatrix},
\end{align}
where $s = \frac{1}{\sqrt{2}}$.
That seems to be a counterexample to your conjecture. So I guess the answer to your question is "Yes, you are just missing something trivial." :) 
