Consider a system described by $q^i(t)$ and its derivatives, by means of a Lagrangian $L=L(q,\dot q)$ and possibly $t$. We say the system is degenerate if $$ \det\left(\frac{\partial L}{\partial \dot q^i\partial\dot q^j}\right)=0 $$ which means that the quadratic form contained in the kinetic term of $L$ cannot be inverted.

On the other hand, we say $L$ has a gauge symmetry if it is invariant under $$ q^i(t)\to q^i(t)+D^i{}_j\lambda^j(t) $$ where $\lambda=\lambda(t)$ is an arbitrary function and $D$ is a certain differential operator.

Question: does degeneracy imply gauge-invariance? what about the converse?

In the case where $L$ is free, $L=\frac12\dot q^i K_{ij} \dot q^j$, with $K$ a differential operator, the answer is straightforward: if $L$ has a gauge symmetry, $K$ has a null-eigenvector and is thus degenerate, and vice-versa. Does a similar analysis hold for more general Lagrangians (i.e., not assuming any particular form for it).