Obviously if you define a first order phase transition as having a discontinuity in the first derivative of the free energy then the answer is no, but I'm asking about if the following situation would still 'effectively' be a first order phase transition in all ways except for the wording of this definition.
The following diagram shows that as you go from a temperature $t$ above the critical temperature $t^*$ to a temperature $t$ below the critical temperature, a previous local minimum of the free energy 'overtakes' the initial global minimum so the system discontinuously jumps to the part of phase space where the new global minimum is - this (I think) is the 'meat' of a first order phase transition.
In the old local minimum, the precise free energy value $\mathcal{L}_0$ at the minimum point will generically continuously vary as the temperature is changed (or in this case (trivially) continuously stay at $\eta=0$. When the system jumps into the new global minimum, this part of the graph will also have a minimum $\mathcal{L}_0$ value which continuosly varies as you vary temperature - but crucially this 'gradient of $\mathcal{L}_0$' is different in the new global minimum than in the old one - this is what causes the cusp in the $\mathcal{L}_0/T$ graph (see second graph below).
My question is, could we envisage a situation where accidentally/coincidentally this new global minimum has a gradient $\frac{d\mathcal{L}_0}{dt}$ which is the same as the old global minimum, therefore not causing a cusp in the $\mathcal{L}_0/T$ graph but nonetheless seeming like a first order phase transition (in that there is a discontinuous jump in the order parameter $\eta$ at the critical temperature (which is indicative of a first not second order transition according to the graph below)?
(I understand this would be unbelievably unlikely on a non-contrived graph)