No, modern standards for symmetric cryptography are heavily over-engineered and the power of chosen plaintext attacks/chosen ciphertext attacks can quickly uncover the structure of such a variant despite its very large key space.
Assuming that you machine is Enigma like with "odometer" stepping, but omitting the Umkehrwalze that introduces the "no fixed point" property, we can describe a set-up device using an initial step position $s$ (which I'll take to be secret rather than a known IV) and fifty permutations $\pi_0,\ldots,\pi_{49}$ on the set of 26 characters. In this case your device would be encrypt the input $x$ at position $t$ in the message as
$$\pi_{49}(\pi_{48}(\pi_{47}(\cdots\pi_1(\pi_0(x+(s+t)\mod{26})+[(s+t)/26]\mod{26})\cdots+[(s+t)/26^{47}]\mod{26})+[(s+t)/26^{48}]\mod{26})+[(s+t)/26^{49}]\mod{26}),$$
or more iteratively/compactly as $y_{50}$ where $y_0=x$ and
$$y_{i+1}=\pi_i\left(y_i+\left[\frac{s+t}{26^i}\right]\right).$$
I'll first describe how to recover $\pi_0$ using CPA analysis. Using Banburismus-like methods, I can recover $s\mod{26}$, I'll take 26 plaintexts where the plaintext characters in the first positions where $(s+t)\mod{26}=0$ and $(s+t)\mod{26}=1$ are AA, BB, CC, DD,... ZZ. As the position of the other rotors does not change between our chosen characters we can treat the encryption of these as $\Pi(\pi_0(x))$ and $\Pi(\pi_0(x+1\mod{26})$ for some fixed permutation $\Pi$. As we have a full set of inputs to these permutations we can remove $\Pi$ by creating a cycle of letters with $x_0=A$ and choosing $x_{i+1}$ such that
$$\Pi(\pi_0(x_i))=\Pi(\pi_0(x_{i+1}+1\mod{26})$$
which by the permutation property tells us that
$$\pi_0(x_i)=\pi_0(x_{i+1}+1\mod{26})$$
which is the tells us the cycle representation of the permutation $\pi_0^{-1}\circ\sigma\circ\pi_0$ where $\sigma$ is the cycle $(0,1,\ldots,25)$. This is a full cycle permutation, of which there are $25!$ so that there are 26 possible values of $\pi_0$ amongst the full $26!$ possible permutations of 26 characters and we should be able to uniquely identify $\pi_0$ in a list of 100.
Once $\pi_0$ is recovered, we can recover $\pi_1$ in a similar fashion by considering chosen plaintexts where $s+t=0\mod{26^2}$ and $s+t=26\mod{26^2}$ and so on for the remaining rotors.
