3-input XNOR using NOR gates only

A 3-input XNOR gives $Y = \overline{A \oplus B \oplus C}$. Use the 4-NOR XNOR block twice and one NOR inverter.

Since $\overline{A \oplus B} \odot C = \overline{\overline{A \oplus B} \oplus C} = A \oplus B \oplus C$, two XNOR blocks give XOR, and a final NOR inverter gives XNOR:

$$\begin{aligned} P &= A \odot B = \overline{A \oplus B} \quad (4 \text{ NOR}) \\ Q &= P \odot C = A \oplus B \oplus C \quad (4 \text{ NOR}) \\ Y &= \overline{Q + Q} = \overline{A \oplus B \oplus C} \quad (1 \text{ NOR}) \end{aligned}$$

Each XNOR block: $G_1 = \overline{X+Z}$, $G_2 = \overline{X+G_1}$, $G_3 = \overline{Z+G_1}$, out $= \overline{G_2+G_3}$.

 A --+
     +--[XNOR: 4 NOR]-- P --+
 B --+                      +--[XNOR: 4 NOR]-- Q
                       C ---+                  |
                 Y --[NOR, inputs tied]--------+

Total: 9 NOR gates.

Y = 1 when an even number of inputs are 1.