If $\sin \left(\frac{\pi}{4} \cot \theta\right)=\cos \left(\frac{\pi}{4} \tan \theta\right)$, then the general solution of $\theta$ is

The equation $\sin \left(\frac{\pi}{4} \cot \theta\right) = \cos \left(\frac{\pi}{4} \tan \theta\right)$ is equivalent to $\sin \left(\frac{\pi}{4} \cot \theta\right) = \sin \left(\frac{\pi}{2} - \frac{\pi}{4} \tan \theta\right)$ using the identity $\cos x = \sin \left(\frac{\pi}{2} - x\right)$.

The general solution for $\sin A = \sin B$ is $A = n\pi + (-1)^n B$, where $n \in \mathbb{Z}$. Applying this yields $\frac{\pi}{4} \cot \theta = n\pi + (-1)^n \left(\frac{\pi}{2} - \frac{\pi}{4} \tan \theta\right)$.

Dividing through by $\frac{\pi}{4}$ gives $\cot \theta = 4n + (-1)^n (2 - \tan \theta)$.

Assuming the arguments of sine and cosine are acute angles, $\sin A = \cos B$ implies $A = \frac{\pi}{2} - B$. Thus, $\frac{\pi}{4} \cot \theta = \frac{\pi}{2} - \frac{\pi}{4} \tan \theta$.

Multiplying both sides by $\frac{4}{\pi}$ yields $\cot \theta = 2 - \tan \theta$, which simplifies to $\cot \theta + \tan \theta = 2$.

Since $\cot \theta + \tan \theta = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin(2\theta)}$, we obtain $\frac{2}{\sin(2\theta)} = 2$, hence $\sin(2\theta) = 1$.

The general solution is $2\theta = 2n\pi + \frac{\pi}{2}$, so $\theta = n\pi + \frac{\pi}{4}$.

This yields $\cot \theta = 1$ and $\tan \theta = 1$, confirming the acute angle assumption and satisfying the original equation.

Final answer: $\boxed{\text{A}}$