Reformulation of the question: Alice is given $n$ ciphers ( $cipher_0,cipher_1,...cipher_n$) to build a key expander, one of these ciphers is a secure cipher (random permutation) and the rest ($n-1$ of them) can be maliciously controlled by Eve. Alice are unable to inspect these ciphers or their input/outputs to determine which one is secure. Alice and Bob both know a key, $k$, they wish to derive a series of generated keys $k'_0, ... k'_n$ from $k$ using the ciphers.
Eve should not be able to figure out $k$ from the generated keys $k'_0, ... k'_n$ without having to break all $n$ of the ciphers.
It seems that Eve can do two malicious things in designing her ciphers (if I'm leaving something out please let me know):
Eve can use the trivial permutation for her ciphers and thereby leak the key. For example: $k = cipher_{eve}(k, k)$.
Eve can reduce the entropy of the output of the cipher. For example if she knows the root of our construction is $cipher_{eve}(k, k)$ where $k$ is the key we are expanding, she can set $cipher_{eve}$ to always return all $0$'s when the plaintext and the key are equal. While this won't allow her to guess the key it might allow her to reduce the entropy of the expanded keys ( $k'_0, ... k'_n$ ).
Consider the case $n = 1$: The construction
$$k'_0 = cipher_0(k, k)$$
is secure. That is, finding the key from $k$ from $k'_0, ... k'_n$ is not possible without guessing $k$ or breaking the cipher, $cipher_0$. Since by our definition $cipher_0$ must be secure, this solution is also secure of $n = 1$.
Consider the case $n = 2$: This case is much more difficult because it introduces a maliciously designed cipher $cipher_{eve}$. The construction
$$k'_0 = cipher_1( cipher_0(k+1, k+1) \oplus k, cipher_0(k, k) )$$
$$k'_1 = cipher_1( cipher_0(k+2, k+2) \oplus k, cipher_0(k, k) )$$
is secure. We can show this by considering both cases:
$cipher_0$ is secure and $cipher_1 = cipher_{eve}$: Because $cipher_1$ doesn't know the value of $k$ it is unable to leak using the trivial permutation. Nor can $cipher_1$ reduce the entropy because the inputs to $cipher_1$ are random and different. Thus if $cipher_0$ is secure this construction is secure.
$cipher_1$ is secure and $cipher_0 = cipher_{eve}$: All outputs from $cipher_0$ will be securely encrypted by $cipher_1$ so $cipher_0$ can't leak the value of $k$. $cipher_0$ can reduce the entropy of it's output ($0^m = cipher_0(k, k)$), but this will not effect the entropy of the input to $cipher_1$ since the entropy of $k$ is preserved by xoring it into the input to $cipher_1$. Thus if $cipher_1$ is secure this construction is secure.
Therefore if $cipher_0$ or $cipher_1$ is secure the secrecy of $k$ is preserved.
The case $n = 2$ also provides the construction for $n > 2$, since the value of $k$ can't be leaked past the secure cipher and the insecure ciphers can't reduce the entropy of the generated keys $k'_0, ... k'_n$.
Does anyone know if this is a known result?
EDIT: I was thinking about this some more and found a serious weakness.
Consider the case in which $cipher_0 = cipher_{eve}$ and $cipher_{eve}$ emits $k$ with probability of $0.5$ when the key is equal to the plaintext.
$$pr( cipher_{eve}(a, b) = k | a = b ) = 0.5$$
Thus half the time $cipher_1$ will get a key of $0$.
Knowing the key that $cipher_1$ gets allows you to recover the original key by decrypting the output of $cipher_1$.
$$k'_0 = cipher_1( k \oplus k, cipher_0(k, k) )$$
$$k'_0 = cipher_1( 0, cipher_0(k, k) )$$
$$cipher_0(k, k) = cipher_1.decrypt(0, k'_0)$$
And since $cipher_0 = cipher_{eve}$ we can invert it half the time and get k.