Short answer: no. Long answer: if you pull the base of any of the transistors in the transistor banks to +5V then any of the signals connected to their emitters will be forced to one Vbe drop below 5V i.e. to about +4.3V. This is because they are pulled up to 5V through the base emitter junction of the transistors. If the logic source can pull down hard enough then it will simply destroy the base emitter junctions. Even if you try to fix this by inserting base resistors or replacing the BJTs with MOSFETs or bidirectional CMOS switches or multiplexers, you may be able to switch USB 1.x data rates but it will struggle to handle the higher USB 2 rates and will completely fail to switch USB 3 signals cleanly. You may be able to find dedicated bidirectional logic switche devices than can handle USB 3 rates but advice on that is outside the scope of what I could offer in the forum. [https://en.wikipedia.org/wiki/USB](https://en.wikipedia.org/wiki/USB)<br> <br> You may be better to look for an off the shelf switching solution. For example: [https://www\.amazon\.co\.uk/Selector\-Computer\-Peripheral\-Switcher\-Keyboard/dp/B08CZH6MQL/ref=asc\_df\_B08CZH6MQL/](https://www.amazon.co.uk/Selector-Computer-Peripheral-Switcher-Keyboard/dp/B08CZH6MQL/ref=asc_df_B08CZH6MQL/)