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If $$U_1 = U_2$$, then $$U_1 U_2^{\dagger} = I$$. So let's add to the first circuit the the inverse/dagger of the second one:

This whole thing should be an identity. Note that, for this, we should prove that whole circuit except last $$R_y(-\theta) I$$ should be equal to $$R_y(\theta) I$$. By remembering that the SWAP gate consists of 3 CNOT gates and $$\text{SWAP} \cdot I R_y(\theta) \cdot\text{SWAP} = R_y(\theta) I$$, we will have:

This should be equal to $$R_y(\theta) I$$. This circuit is equal to $$UR_y(\theta) I U^\dagger$$$$U \cdot R_y(\theta) I \cdot U^\dagger$$, where $$U$$ is this circuit:

Now we should try to prove that $$UR_y(\theta) I U^\dagger = R_y(\theta) I$$$$U \cdot R_y(\theta) I \cdot U^\dagger = R_y(\theta) I$$. After simplifying the $$U$$ I have obtained this circuit:

where I took into account that $$HXH = Z$$ and $$I \otimes H \cdot \text{CNOT} \cdot I \otimes H = CZ$$ (note that I have omitted the $$\otimes$$ sign in the previous expressions). If my calculations are right, then $$U = \frac{1}{\sqrt{2}}(I \otimes I+iY \otimes X)$$. So, by omitting again the $$\otimes$$ sign:

$$UR_y(\theta) I U^\dagger = \frac{1}{2}\left[II+iY X\right]\left[\cos(\theta)I I - i\sin(\theta)Y I\right]\left[I I-iY X\right] = R_y(\theta) I$$$$U \cdot R_y(\theta) I \cdot U^\dagger = \frac{1}{2}\left[II+iY X\right]\left[\cos(\theta)I I - i\sin(\theta)Y I\right]\left[I I-iY X\right] = R_y(\theta) I$$

So, the circuits are equivalent.

deleted 130 characters in body

If $$U_1 = U_2$$, then $$U_1 U_2^{\dagger} = I$$. So let's add to the first circuit the the inverse/dagger of the second one:

This whole thing should be an identity. Note that, for this, we should prove that whole circuit except last $$R_y(-\theta) I$$ should be equal to $$R_y(\theta) I$$. By remembering that the SWAP gate consists of 3 CNOT gates and $$\text{SWAP} \cdot I R_y(\theta) \cdot\text{SWAP} = R_y(\theta) I$$, we will have:

This should be equal to $$R_y(\theta) I$$. This circuit is equal to $$UR_y(\theta) I U^\dagger$$, where $$U$$ is this circuit:

Now we should try to prove that $$UR_y(\theta) I U^\dagger = R_y(\theta) I$$. After simplifying the $$U$$ I have obtained this circuit:

where I took into account that $$HXH = Z$$ and $$I \otimes H \cdot \text{CNOT} \cdot I \otimes H = CZ$$ (note that I have omitted the $$\otimes$$ sign in the previous expressions). If my calculations are right, then $$U = \frac{1}{\sqrt{2}}(I \otimes I+iY \otimes X)$$. So, by omitting again the $$\otimes$$ sign:

$$UR_y(\theta) I U^\dagger = \frac{1}{2}\left[II+iY X\right]\left[\cos(\theta)I I - i\sin(\theta)Y I\right]\left[I I-iY X\right] = R_y(\theta) I$$

So, the circuits are equivalent.

If $$U_1 = U_2$$, then $$U_1 U_2^{\dagger} = I$$. So let's add to the first circuit the the inverse/dagger of the second one:

This whole thing should be an identity. Note that, for this, we should prove that whole circuit except last $$R_y(-\theta) I$$ should be equal to $$R_y(\theta) I$$. By remembering that the SWAP gate consists of 3 CNOT gates and $$\text{SWAP} \cdot I R_y(\theta) \cdot\text{SWAP} = R_y(\theta) I$$, we will have:

This should be equal to $$R_y(\theta) I$$. This circuit is equal to $$UR_y(\theta) I U^\dagger$$, where $$U$$ is this circuit:

Now we should try to prove that $$UR_y(\theta) I U^\dagger = R_y(\theta) I$$. After simplifying the $$U$$ I have obtained this circuit:

where I took into account that $$HXH = Z$$ and $$I \otimes H \cdot \text{CNOT} \cdot I \otimes H = CZ$$ (note that I have omitted the $$\otimes$$ sign in the previous expressions). If my calculations are right, then $$U = \frac{1}{\sqrt{2}}(I \otimes I+iY \otimes X)$$. So, by omitting again the $$\otimes$$ sign:

$$UR_y(\theta) I U^\dagger = \frac{1}{2}\left[II+iY X\right]\left[\cos(\theta)I I - i\sin(\theta)Y I\right]\left[I I-iY X\right] = R_y(\theta) I$$

So, the circuits are equivalent.