For invertible matrix $A$, we have
$\log(\det A) = \mathrm{tr}(\log A)$
due to a corollary of Jacobi's formula. What if we had the argument $iA$ instead? Would the above relation still hold?
Edit:
RHS $= \mathrm{tr}(\log(iA)) = \mathrm{tr}(i \tfrac{\pi}{2}+\log A) = i \tfrac{\pi}{2} + \mathrm{tr}(\log A) = i \tfrac{\pi}{2} + \log(\det A)$
$$\mathrm{tr}(\log (iA )) = \mathrm{tr}(i \pi/2 + \log A ) = i n \pi/2 + \mathrm{tr}(\log A )$$ When you multiply a matrix by a constant $c$, the determinant gets a factor of $c^n$. $$\log(\det ( i A )) = \log( i^n \det A ) = \log( i^n ) + \log( \det A ) = \log( i^n ) + \mathrm{tr}\log( A )$$
So in your case for the RHS to equal the LHS you want
$$\log(i^n) = i n\pi/2$$
But $$\log(i^n) = \{ 0, i \pi/2, i\pi, i 3 \pi/2 \}$$ for $n$ modulo 4.
This is only going to work for you if $n < 4$. Otherwise it is false.
EDIT: In a sense it is true if you are going to treat the output as an angle for some complex number. In other words your equality is still true modulo $2 \pi$.