oversampled coefficient for existing exponential smoothing
I think it's instructive to take a look at the impulse responses, taking absolute time into account. I'm assuming the following recursion:

$$y_n=\alpha x_n+(1-\alpha)y_{n-1}\tag{1}$$

The impulse response corresponding to the recursive system described by (1) is

$$h_n=\alpha(1-\alpha)^n$$

Assume now that we have a discrete-time system with sample interval $T$, and another one with sample interval $T^{\prime}=T/L$. Let $\beta=1-\alpha$, and let $\alpha^{\prime}$ and $\beta^{\prime}$ be the constants of the system with sample interval $T^{\prime}$. In order for the two systems to have the same time constant and the same scaling we require (considering a point in time $t_n=nT=nLT^{\prime}$)

$$\alpha\beta^{n}=\alpha^{\prime}{\beta^{\prime}}^{nL}\tag{2}$$

From (2) we see that

$$\alpha=\alpha^{\prime},\quad \beta=\beta^{\prime L}\quad
\text{(i.e. $\beta^{\prime}\neq 1-\alpha^{\prime}$)}$$

In this case the impulse response of the system with the higher sampling rate is simply an interpolated version of the other system's impulse response. Note that the condition on $\beta^{\prime}$ is the one that you came up with intuitively, and it is also the more important of the two because it affects the time constant of the system. The choice $\alpha=\alpha^{\prime}$ only affects the scaling and in this sense is secondary. Choosing $\alpha^{\prime}=1-\beta^{\prime}$ is of course also possible and just changes the scaling.