Let’s define a transducer as a $5$-tuple $(Q, A, B, \phi, \psi)$, where $Q$ is the collection of states, $A$ is the input alphabet, $\phi: Q\times A \to Q$ is the transition function and $\psi: Q \times A \to B$ is the output function.
Any transducer defines a transducer function $\overline{\psi}: Q\times A^* \to B^*$ described by the following recurrence:
$\overline{\psi}(q, \Lambda) = \Lambda$, where $\Lambda$ is the empty word.
$\overline{\psi}(q, a \alpha) = \psi(q, a) \overline{\psi}(\phi(q, a), \alpha)$, where $a \in A$, $\alpha \in A^*$.
We call a transducer invertible iff $(q, a) \mapsto (\phi(q, a), \psi(q, a))$ is a bijection between $Q \times A$ and $Q \times B$. Note, that a transducer can only be invertible when $|A| = |B|$. It is also not hard to see, that if a transducer is invertible, then $\forall q \in Q$ $\alpha \mapsto \overline\psi(q, \alpha)$ is a bijection between $A^*$ and $B^*$.
Now, suppose $V = (A, Q, A, \phi, \psi)$ is an invertible transducer in which output alphabet and input alphabet are the same. Then $\sigma_q: \alpha \mapsto \overline{\psi}(q, \alpha)$ is an element of $Sym(A^*)$. Let's define a transducer group $\langle V \rangle = \langle \{\sigma_q |q \in Q\}\rangle<Sym(A^*)$.
It is not hard to see, that for any $k$-generated group $G$ of order $n$, there exists a $k$-state invertible transducer $V$ over the $n!$ letter alphabet, such that $G \cong \langle V \rangle$. Indeed, suppose $G = \langle S \rangle$. Then we can take a transducer $V=(A, Q, B, \phi, \psi)$ in which $A = B = G$, $Q = S$, $\phi: (q, g) \mapsto q$ and $\psi: (q, g) \mapsto qg$. Then $V$ will be invertible and $\langle L \rangle \cong G$.
However, this "transducer representation" quite unlikely to be the best possible one. My question is:
Suppose $t(n, k)$ is the minimal number $m$, such that any $k$-generated group of order $n$ is isomorphic to the transducer group of some $k$-state invertible transducer over $m$-letter alphabet. Is there an explicit formula (or at least asymptotics) for $t(n, k)$?
The only thing I currently know, is that $t(n, k) \leq n!$.