Let $\Lambda$ be a lattice in $\mathbb{C}$ and $\sigma\in\mathrm{Aut}(\mathbb{C})$. Is it true that \begin{equation} \tag{$*$} \sigma(\wp(z; \Lambda))=\wp(\sigma(z); \sigma(\Lambda))? \end{equation} Of course, this is obvious if $\sigma$ is continuous, but most automorphisms of $\mathbb{C}$ are not continuous.
However, I still suspect this to be true for the following reason: If $E/\mathbb{C}$ is an elliptic curve such that $E(\mathbb{C})\cong\mathbb{C}/\Lambda$ and $[\,\cdot\,]_E$ denotes the normalised isomorphism from $\mathcal{O}\subseteq\mathbb{C}$ to $\mathrm{End}(E)$, then $\sigma([\alpha]_E)=[\sigma(\alpha)]_{\sigma(E)}$ for any $\alpha\in\mathcal{O}$ (see Silverman, for example). Chasing this around a few diagrams then seems to imply $(*)$.
So I guess I am really asking two things here: First, is $(*)$ indeed true and second, if so, can we see this directly, i.e. without the detour using elliptic curves?
This is not true. For instance, take $\Lambda$ to be generated by $1$ and $i$, so $\sigma(\Lambda)$ is always equal to $\Lambda$. Then you are asking whether the $\wp$-function satisfies $\sigma(\wp(z))=\wp(\sigma(z))$. Now take any $z\in\mathbb{C}$ such that $z$ and $\wp(z)$ are algebraically independent over $\mathbb{Q}$. (Such a $z$ exists by the Baire category theorem, for instance: for any nonzero $p\in \mathbb{Q}[x,y]$, the set of $z$ such that $p(z,\wp(z))=0$ is closed and has empty interior, and there are only countably many such $p$.) An automorphism of $\mathbb{C}$ can then send $z$ and $\wp(z)$ to any other pair $a,b$ of algebraically independent elements of $\mathbb{C}$. In particular, you can choose such a pair for which $b\neq \wp(a)$, so $\sigma(\wp(z))\neq\wp(\sigma(z))$.