Phase Invterter Circuit
I think you are using the classical IV converter:







simulate this circuit &ndash; Schematic created using CircuitLab

Where:

$$V_u = -I_{in}R_1$$

that's why your waveform is inverted. There are a couple of ways to circumvent that.

You can add an inverting unity gain amplifier after your current stage, something like this:







simulate this circuit

Now:

$$
V_u = R_1 I_{in}
$$

If you want to use a single op amp there is another possibility. Please note that this second circuit could not work with your setup.







simulate this circuit

Now:
$$V_u = R_1 I_{in}$$

The (big) problem with the last circuit is that your current will see a varying voltage so your sensor must be capable to provide it. For example, if you have \$I_{in}=1mA\$ and you choose \$R_1=1k\Omega\$ the voltage at the positive input of the opamp will be 1V. This might or might not be ok for your sensor depending on how it's made. The first two schematics given instead keep the voltage seen by the sensor (very) near to the reference tied to the non inverting input of the op amp, being nicer to the sensor output circuitry.