Cherry-picking the answer from YCor's and Colin Reid's comments:

If $G$ has order $n$, then $|\Delta| \leq \log_2 n + 1$. This bound is sharp in the sense that it is attained for a sequence of groups of unbounded order (namely elementary abelian 2-groups).

Proof:

Let $G$ be a finite group and let $\Delta$ be a generating set with the property described in the OP (i.e. any permutation of $\Delta$ induces an automorphism of $G$).

Let $k$ be the cardinality of $\Delta$ and let $\Delta = \{g_1,\dots,g_k\}$. (Note this fixes a total order on $\Delta$.)

For any $j=1,\dots,k$, let $\Delta_j := \{g_1,\dots,g_j\}$ and let $G_j$ be the subgroup generated by $\Delta_j$. For $j\leq k-2$, $G_j$ must generate a proper subgroup of $G_{j+1}$, because $G$ has automorphisms that fix $G_j$ pointwise but do not fix $G_{j+1}$ pointwise (namely, those induced by any permutation of $\Delta$ that fixes $\Delta_j$ pointwise but moves $g_{j+1}$; note that for $j\leq k-2$, these exist).

As a consequence, $[G_{j+1}:G_j]\geq 2$ for $j\leq k-2$.

Also note $|G_1|\geq 2$ since $g_1$ cannot be the identity as there exist automorphisms of $G$ that do not fix it (as long as $|\Delta|\geq 2$).

Thus, by induction, $|G_j|\geq 2^j$ for $j$ up to $k-1= (k-2)+1$. In particular, $G\supset G_{k-1}$ must have order at least $2^{k-1}$.

This yields the bound given above.

This bound is attained for elementary abelian 2-groups of order $n=2^{k-1}$, for $k\geq 3$: let $\Delta$ consist of a basis (of $k-1$ elements) plus the product of the basis elements. (The condition $k\geq 3$ guarantees the product is distinct from any of the basis elements.) This choice of generators satisfies the condition $\prod g_i = 1$ inside the group, thus the basis elements $g_1,\dots,g_{k-1}$ can be sent to any $k-1$ of the $k$ elements of $\Delta$, inducing an automorphism, and the last element of $\Delta$ will automatically end up in the right place.