You are right. You can take the 10-th root of the odds ratio (relative change of the odds) for a step of size 1, in order to get the odds ratio for a step of size 0.1
The reason is because a logistic model is modeling the logarithm of the odds as a linear function of the parameter/predictor. And this makes the odds an exponential function of the parameter/predictor.
$$\begin{array}{rrcl}\text{log-odds:} \qquad \qquad & \log\left( \frac{p}{1-p} \right) &=&a+bx\\
\text{odds:} \qquad \qquad & \frac{p}{1-p} &=&e^{a+bx}
\end{array}$$
So the changes in the odds are multiplicative when we are adding multiple effects (or increasing the size of an effect).
Consider a change of the parameter/predictor by some step $x_\Delta$. Then this is equivalent to the odds being multiplied by a factor $e^{bx_\Delta}$
$$e^{a+b(x+x_\Delta)} = e^{a+bx} \cdot e^{bx_\Delta}$$
and adding more of those steps (or only a fraction of that step) equates to more multiplications with that factor
$$e^{a+b(x+kx_\Delta)} = e^{a+bx} \cdot e^{bkx_\Delta} = e^{a+bx} \cdot \left(e^{bx_\Delta}\right)^k$$
The image below shows the different representations (probability, odds and log-odds) as function of the parameter $x$. The logistic model $p = (1+\exp(x))^{-1}$ will model the log-odds as a linear function of $x$.
