$\DeclareMathOperator{\Sym}{Sym}\DeclareMathOperator{\Spec}{Spec}$ Let $S$ be a smooth variety over an algebraically closed field $k$, $\operatorname{char}k \neq 2$, and let $\Sym^2(S) = (S \times_k S) / (\mathbb Z/2)$ be the second symmetric power of $S$. Then $S$ embeds diagonally as a closed subscheme into $\Sym^2(S)$, $$\delta: S \to \Sym^2(S).$$ Clearly $\delta$ is the composition of the ordinary diagonal $\Delta: S \to S \times S$ with the quotient map $p: S \times S \to \Sym^2(S)$.
I would like to understand the conormal sheaf $\mathcal C_{S / \Sym^2(S)} = I_\delta / I_\delta^2$. My first suspicion that $$p^* I_\delta = I_\Delta^2$$ is actually not correct, since the group quotient is not flat.
See my answer below, I was still able to show $$\mathcal C_{S / \Sym^2(S)} = \delta^* I_\delta = \Delta^* p^* I_\delta = \Delta^* I_\Delta^2 = I_\Delta^2 / I_\Delta^3.$$
As $p^{-1}(\delta(S)) \subset \Delta(S)$, we have a morphism of sheaves $$p^* I_\delta \to I_\Delta \subset \mathcal O_{S \times S}.$$ Claim. This morphism gives a surjection $p^* I_\delta \twoheadrightarrow I_\Delta^2$, which becomes an isomorphism $\delta^* I_\delta \cong I_\Delta^2 / I_\Delta^3$ after pulling-back via $\Delta$.
Proof. We first deal with the case $S = \mathbb A^n = \operatorname{Spec}(k[s_1, \dotsc, s_n])$, where this can be done by explicit calculation. On $S \times S$ we have then coordinates $s_1, \dotsc, s_n, t_1, \dotsc, t_n$ and the $\mathbb Z/2$-action is $\sigma(s_i) = t_i$. Introducting new coordinates $$x_i = \frac 1 2 (s_i + t_i) \quad \text{and} \quad \xi_i = \frac 1 2 (s_i - t_i)$$ we see that the subring of invariants is $$R^{\mathbb Z /2} = k[x_1, \dotsc, x_n, \xi_1^2, \xi_1 \xi_2, \dotsc, \xi_1 \xi_n, \xi_2^2, \xi_2 \xi_3, \dotsc] \subset k[x_1, \dotsc, x_n, \xi_1, \dotsc, \xi_n] = R.$$ Then $I_\Delta = (\xi_1, \dotsc, \xi_n) \subset R$, and $I_\delta = (\xi_i \cdot \xi_j \,|\, i,j= 1, \dotsc, n) \subset R^{\mathbb Z/2}$. So $I_\delta$ generates the ideal $I_\Delta^2$ over $R$, which means we have the surjection $I_\delta \otimes R \twoheadrightarrow I_\Delta^2$. Note that this is not injective (the group quotient is not flat). For example elements of type $\xi_i \xi_j \otimes \xi_k - \xi_j \xi_k \otimes \xi_i$ are in the kernel. However, if we mod out $(\xi_1, \dotsc, \xi_n)$ this becomes injective: The kernel of the composite map $I_\delta \otimes R \to I_\Delta^2 \to I_\Delta^2 / I_\Delta^3$ is exactly $$I_\delta \otimes I_\Delta \subset I_\delta \otimes R,$$ which is mapped to zero under the quotient $I_\delta \otimes R \to I_\delta \otimes R/I_\Delta$. So we obtain an isomorphism $$\delta^* I_\delta = I_\delta \otimes R/I_\Delta \cong I_\Delta^2 / I_\Delta ^3.$$ Now to the general case. As the map $p^*I_\delta \to I_\Delta$ is defined globally, it is sufficient to check the claim on a Zariski cover. By Stacks 054L we may assume that $S$ is affine, and admits an étale morphism $f: S \to \mathbb A^n$. Then $f$ induces morphisms $S \times S \to \mathbb A^n \times \mathbb A^n$ and $\operatorname{Sym}^2(S) \to \operatorname{Sym}^2(\mathbb A^n)$, leading to a commutative diagram $$\require{AMScd} \begin{CD} S \times S @>f \times f>\text{étale}> \mathbb A^n \times \mathbb A^n\\ @VpVV @VVV \\ \operatorname{Sym}^2(S) @>{}>> \operatorname{Sym}^2(\mathbb A^n) \\ @AAA @AAA \\ S @>{\text{étale}}>f> \mathbb A^n \end{CD}$$ Note however, that the middle horizontal arrow is in general not étale. If you have two points $s \neq t \in S$ with $f(s) = f(t)$, the corresponding point $[s+t] \in \operatorname{Sym}^2(S)$ is a smooth point. But it maps to the singular point $[2f(s)] \in \operatorname{Sym}^2(\mathbb A^n)$.
Since $f \times f$ is étale, the preimage of $\Delta(\mathbb A^n)$ decomposes as a disjoint union $(f \times f)^{-1}(\Delta(\mathbb A^n)) = \Delta(S) \sqcup Z$, and $\operatorname{Sym}^2(S) \setminus p(Z)$ is an open neighbourhood of $\delta(S)$, mapping étale to $\operatorname{Sym}^2(\mathbb A^n)$. So the following diagrams are cartesian, all horizontal morphisms étale: $$ \begin{CD} S @>f>> \mathbb A^n @. S @>f>> \mathbb A^n\\ @V\delta_SVV @VV\delta_{\mathbb A^n}V @VVV @VVV\\ \operatorname{Sym}^2(S) \setminus p(Z) @>>g> \operatorname{Sym}^2(\mathbb A^n) @. S \times S \setminus Z @>>> \mathbb A^n \times \mathbb A^n \end{CD}$$ From this and the flatness of $g$ we obtain $$\delta_S^* I_{\delta_S} = \delta_S^* g^* I_{\delta_{\mathbb A^n}} = f^* \delta_{\mathbb A^n}^* I_{\delta_{\mathbb A^n}} = f^* I_{\Delta_{\mathbb A^n}}^2 / I_{\Delta_{\mathbb A^n}}^3 = I_{\Delta_S}^2 / I_{\Delta_S}^3.$$