Level shift 0/5V to -10V/+10V
Opamps are the universal answer to everything today. They are reasonably priced, precise, widely available, can provide rail-to-rail input and output, etc. But since you added "some transistor" and since MOSFETs are more pricey, by and large, than BJTs and have their own unique foibles, I'll give a simple BJT solution that is really easy to understand.

I've added no "speed-ups" to it, since your specification says "under \$10\:\textrm{Hz}\$." I've also avoided adding reverse voltage \$V_{BE}\$ protection diodes for any of the transistors. I don't think that will be an issue for you, more especially because I didn't add any speed-up caps. I'm going to also assume (it's reasonable, given the question) that you have the following rails: \$+5\:\textrm{V}\$, \$+10\:\textrm{V}\$, \$-10\:\textrm{V}\$, and a common ground (obviously) that they all share.

Finally I'm going to make some assumptions: (1) that your source can handle \$\ge 200\:\mu\textrm{A}\$, both source and sink; and, (2) that you can accept an output that won't need more than 20-30 times that, or so. (This depends upon the actual output BJTs you use and just how "saturated" you want them.) So you should NOT use a load of more than about \$2\:\textrm{k}\Omega\$ when using the exact values I'm providing. If your load will be heavier than this by some factor \$X\$, you'll need to reduce the values for \$R_1\$ and \$R_2\$ by that same factor \$X\$ and your input source will need to be able to sink and source \$X\$ times what I mentioned above, as well. (Oh, and you can remove \$R_5\$ and \$R_6\$, if you want. They are not essential. If you keep them, also adjust them by the same factor \$X\$ you used to adjust \$R_1\$ and \$R_2\$.) The values for \$R_3\$ and \$R_4\$ aren't critical. You can leave them unchanged, or change them, accordingly.

I'm using somewhat higher voltage BJTs for the output. I just figured I'd make this capable of much higher rail voltages, in case anyone cares. Feel free to replace them as you see fit (with small signal BJTs like the 2n3904 and 2n3906, if you want.)

Feel free to jump to the bottom, where probably the better answer is located. I present a few steps in getting there. But if you are in a rush, just skip all of it and go there.

Now to get started with a direct approach:







simulate this circuit &ndash; Schematic created using CircuitLab

Above, I completely ignored your suggestion that \$0\:\textrm{V}\$ on the input generates \$-10\:\textrm{V}\$ on the output. Instead, the above circuit will put \$+10\:\textrm{V}\$ on the output with \$0\:\textrm{V}\$ on the input and will put \$-10\:\textrm{V}\$ on the output with \$+5\:\textrm{V}\$ on the input.

But it's easy to explain. The input source either "pulls down" on the emitter of \$Q_3\$, which sinks base current of \$Q_1\$, turning \$Q_1\$ on; or else it "pulls up" on the emitter of \$Q_4\$, which sources base current into \$Q_2\$, turning \$Q_2\$ on. That's really about all there is to it.



If you don't like the polarity, though, you can place an inverter at the input, like this:







simulate this circuit

Kind of brute-force, but it gets the job done. But if you need more drive capability in the original circuit, then the added inverter in this latest circuit will also need to be modified. Just a warning.

I'm not going to spend any more time on this one except to say that it points up how you might modify someone else's circuit to get the polarity you wanted. On to the next step.



To avoid the inverter and to arrange things to get the output polarity you asked for, you can try this:







simulate this circuit

This is actually probably better still and it provides the polarity you asked about. I've added a series resistor, \$R_7\$, just in case there is any concern about oscillation around the transition edges.

It works in a similar way, though now the input source directly drives the bases rather than pulling on the emitters. But it also doesn't load down your input source nearly as much. So that's why it's probably better for typical uses.

But there's a small problem with it, too. Before I talk about that, a short note:



Just be aware that the above designs are not meant to be operated linearly. A square wave input will be fine, because the transitions are fast. (Sure, I've added some emitter degeneration just in case. But it's still not a good idea to operate them with slow rise and fall times.)



Okay. Now back to the small problem.

The values for \$R_3\$ and \$R_4\$ are weak in the above diagrams. They perform several functions (cheaply), but one of them is to help with the transitions.

There's a base-collector capacitance for all BJTs, but in the case of this topology the base-collector capacitance for \$Q_3\$ and \$Q_4\$ are the problem. When the input pulls one way (or another) at the base of one of these two BJTs in order to turn it off, then this small capacitance also pulls similarly on the bases of \$Q_1\$ and \$Q_2\$ causing them to momentarily increase their collector currents just prior to turn-off; right at the same moment when the opposing transistor (\$Q_2\$ or \$Q_1\$) is turning on. This is not so good, as it connects the power rails together through the two output BJTs for a short moment and that wrecks some havoc. (It might only be for a couple of microseconds. But it's worth seeing if it can be reduced or removed.)

If you can afford the extra currents in the driver BJT stages, you could simply reduce the values for \$R_3\$ and \$R_4\$ to strengthen them a bit to supply the needed transition currents more readily. This reduces the time period and therefore the peak "short-circuit" currents, lost power, and attending havoc. But I set them very weak here, as it doesn't matter as much as one might hope. Still, you could cut them down by a factor of 10. But the problem, while smaller now, would still be present.

Another (better, I think) approach would be to instead add one small capacitor as shown below:







simulate this circuit

This works because just about the only node moving upward in voltage when the input is driving downward, is the base of \$Q_2\$. (The reverse concept is also true.) By tying a capacitor between those two nodes, the turn-on of one side aids the turn-off of the other side. This compensates for the base-emitter capacitance on both sides and nearly eliminates the spikes during the transitions.

The value of the capacitor needs to be sufficient to overwhelm the base-collector capacitance of \$Q_3\$ and \$Q_4\$ (which are different for various BJTs) and the value I've added is large enough to handle several different BJTs you are likely to encounter. But it could be still larger as it's not critical as it has access to substantial currents.