It is a basic fact of set theory that the following holds:

Let $(X,\leq)$ be a linear ordered set. Then it holds that, for each finite sequence $y_{1}$,...,$y_{n}$ of sets and each formula $\phi(x,y_{1},...,y_{n})$ of set theory, we have

$\forall{y_{1},...,y_{n}}$ $(\forall{x\in X}(\forall{z}$<$x\phi(z,y_{1},...,y_{n})\rightarrow\phi(x,y_{1},...,y_{n}))\rightarrow\forall{x\in X}\phi(x,y_{1},...,y_{n}))$

iff $(X,\leq)$ is a well-ordering. Hence induction works on $(X,\leq)$ iff $(X,\leq)$ is a well-ordering.

We can ask the same question for the recursion principle:

Let $(X,\leq)$ be a linear ordering such that, for each function $F:V\times V\times V\rightarrow V$, and each parameter $p$, there is a unique function $f:X\rightarrow V$ such that $f(x)=F(x,f|X_{x},p)$ for all $x\in X$, where $X_{x}$ is the subset of $X$ consisting of all elements strictly smaller than $x$ and $f|A$ is the restriction of $f$ to $A$.

Does it then follow that $(X,\leq)$ is a well-ordering?

This is obviously true if we assume the axiom of foundation. E.g. we could otherwise use $F(x,y)=y$ to construct an ill-founded set.

However, the first fact about induction seems to be more 'logical' in nature and does not depend on the axiom of foundation. I would therefore be interested in knowing whether the second fact can also be deduced without assuming foundation.